8  4  ess 


•* 


m 


UNIVERSITY  OF  CALIFORNIA. 


Class 


HINTS  FOR  CRYSTAL   DRAWING 


HINTS 


FOR 


CRYSTAL    DRAWING 


BY 


MARGARET  REEKS 
\\ 


WITH  A   PREFACE  BY 

JOHN  W.  EVANS,  D.Sc.,  LL.B.,   F.G.S. 

MINERALOGIST   TO  THE   SCIENTIFIC   AND  TECHNICAL  DEPARTMENT   OF  THE 
IMPERIAL   INSTITUTE 


WITH  ILLUSTRATIONS  BY  THE  AUTHOR 


LONGMANS,    GREEN,    AND    CO. 

39   PATERNOSTER   ROW,   LONDON 

NEW   YORK,   BOMBAY,  AND   CALCUTTA 

1908 


GENERAL 


TO 

PROFESSOR  J.  W.  JUDB,  C.B.,  LL.D.,  F.K.S., 

LATE   DEAN  AND  PROFESSOR  OP  GEOLOGY  AT  THE  ROYAL  COLLEGE  OF  SCIENCE 
WITH    SINCEBE   RESPECT 


197729 


OF   THE 

UNIVERSITY 

OF 


PEEFACE. 

THE  accurate  representation  of.  geometrical  re- 
lations in  three  dimensions  is  always  a  matter  of 
difficulty,  and  this  is  especially  the  case  with  the 
forms  of  crystals.  The  most  obvious  course  is  to 
resort  to  models,  but  these  are  troublesome  to  con- 
struct, and  occupy  considerable  space,  so  that  for 
all  practical  purposes  some  method  of  representa- 
tion in  two  dimensions,  in  other  words  a  projection 
on  a  plane  surface,  such  as  a  sheet  of  paper,  must 
be  employed. 

The  stereographic,  gnomonic  and  linear  projec- 
tions, though  admirable  as  exact  records  of  the 
disposition  in  space  of  the  faces,  are  unsuited  to 
the  elementary  student  who  requires  a  projection 
that  will  enable  him  to  realise  at  once  the  general 
form  of  the  crystal.  Ordinary  perspective  is,  on 
the  other  hand,  objectionable  because  the  parallel- 
ism of  lines  is  not  preserved,  and  the  scale  of 
representation  varies  according  to  the  distance  from 
the  point  of  view.  It  therefore  becomes  necessary 
to  substitute  parallel  lines  of  sight  or  projection 
for  those  converging  to  the  observer's  eye,  or  to 
use  a  convenient,  if  somewhat  self-contradictory 

vii 


viii  PEEFACE 

locution,   to   represent   the    object  as   seen  from 
infinity. 

There  still,  however,   remains  a  considerable 
choice  as  to  the  procedure  to  be  adopted.    We 
may,  if  we  like,  construct  a  plan  and  front  and 
side  elevations — in  other  words,  representations  of 
the   object  on  a  horizontal  plane,  as  seen   from 
above,   and   on   vertical    planes,   as   seen   from   a 
horizontal  direction.     In  these,  the  lines  of  pro- 
jection are  at  right  angles  to  the   plane  of  the 
drawing   or   projection,   so   that   this   method    of 
representation  is  a  special  case  of  "  orthographic 
projection".      It  is  of  great  value  for  both  edu- 
cational and  purely  scientific  purposes,  but  has  the 
drawback  that,  as  a  rule,  less  than  half  of  the 
faces  are  seen  as  such  in  the  same  view,  others 
being  represented  only  by  their  edges,  which  form 
the  whole  or  part  of  the  outline  of  the  figure.     A 
better  idea  of  the  crystal  is  therefore  obtained  if 
it  be  viewed  obliquely  from  above  and  one  side. 

It  remains  to  consider  the  position  of  the  plane 
of  projection,  the  canvas  of  our  picture.  Formerly 
it  was  usually  placed  at  right  angles  to  the  lines  of 
sight — another  example  of  orthographic  projection. 
It  is  now,  however,  considered  preferable  to  follow 
the  usages  of  pictorial  art  and,  while  placing  the 
plane  of  representation  or  projection  facing  the 
supposed  direction  of  the  observer,  to  keep  it  in 
an  upright  position.  Every  photographer  knows 
the  importance  of  keeping  the  plate-holder  upright 


PEEFACE  ix 

if  a  natural  picture  is  to  be  obtained.  The  descend- 
ing lines  of  sight  will  then  meet  the  plane  at  an 
oblique  angle,  and  the  projection  is  therefore  re- 
ferred to  as  inclined  or  clinographic.  The  crystal 
will  accordingly  stand  with  its  vertical  axis  parallel 
to  the  plane  of  projection,  but  its  principal  hori- 
zontal directions  will  be  oblique  to  the  observer, 
and  therefore  to  the  plane  of  projection ;  so  that 
the  crystal  may  be  considered  as  rotated  on  its 
vertical  axis  relatively  to  both. 

As  a  result  of  these  relations  vertical  directions 
retain  their  original  dimensions ;  but  distances 
from  front  to  back  or  right  to  left  are  diminished 
in  certain  fixed  proportions  dependent  upon  the 
angle  of  rotation  and  the  angular  elevation  of  the 
line  of  sight. 

In  the  present  work  the  author,  who  has  had  a 
long  and  varied  experience  in  mathematical  draw- 
ing of  every  description,  has  devoted  herself  with 
her  usual  energy  to  the  drawing  of  crystals  in 
clinographic  projection.  While  following  in  many 
respects  the  method  of  the  late  Professor  Pen- 
field,  she  has  been  able  to  give  the  student  the 
advantage  of  numerous  practical  expedients  which 
will  greatly  facilitate  his  work,  and  in  some  cases 
she  has  made  distinct  contributions  to  mathematical 
draftsmanship.  Her  method  of  determining  the 
position  of  the  axes  in  the  triclinic  system  is,  in 
fact,  an  ingenious  graphic  solution  of  a  spherical 
triangle  when  three  sides  are  given.  Her  pro- 


x  PEEFACE 

cedure  in  drawing  twin  crystals  is  also  marked  by 
simplicity  and  originality. 

If  the  student  who  is  commencing  the  study 
of  the  geometry  of  crystals  will  work  carefully 
through  at  least  the  simpler  of  the  author's  con- 
structions, and  follow  this  up  by  drawing  other 
examples  by  the  same  methods,  he  will  obtain  a 
clearer  grasp  of  crystal  forms  than  he  could  possibly 
do  by  means  of  mere  book  work.  Even  the  value 
of  models  for  purposes  of  instruction  is  seriously 
discounted  without  the  constructive  training  which 
is  afforded  by  a  course  of  crystal  drawing. 

JOHN  W.  EVANS. 


IMPERIAL  INSTITUTE, 
8th  April,  1908. 


AUTHOR'S   PREFACE. 

BY  way  of  preface,  I  would  ask  one  boon  of  the 
teacher  of  Crystallography  on  behalf  of  the  student, 
already  bewildered  by  the  difficulties  of  the  sub- 
ject— that  he  should  not  be  confronted  by  a 
needless  complexity  of  nomenclature  and  nota- 
tion, but  that  one  scheme  should  be  selected  and 
authorised  by  use. 

The  classification  and  symbols  employed  in  this 
book  have  been  very  generally  adopted.  They 
are  used  in  the  classes  at  the  "  Imperial  College 
of  Science  and  Technology,"  and  in  many  of  the 
standard  text-books. 

So  admirably  has  our  great  national  collection 
of  minerals,  at  the  Natural  History  Museum, 
South  Kensington,  been  arranged,  that  the  student 
who  has  no  opportunity  of  attending  classes,  may 
there  see  in  the  cases  the  entire  science  of  miner- 
alogy, as  it  were  written  in  minerals  themselves. 

I  have  only  to  add,  with  regard  to  the  solutions 
given  in  this  book,  that  it  is  quite  possible,  that 
preferable  methods  may  present  themselves  to  the 
individual  student  for  any  particular  point  of  con- 
struction— tant  mieux,  let  him  follow  his  own  road 


xii  AUTHOE'S  PEEFACE 

— nevertheless  it  is  hoped  that  the  hints  supplied 
will  aid  him  when  struggling  with  the  irritating 
elusiveness  of  crystal  form. 

It  is  believed  that  all  axial  ratios  and  other 
data  given  are  accurate,  but  should  any  error  have 
crept  in,  it  will  not  affect  the  demonstration. 

In  conclusion,  I  would  gratefully  thank  Dr. 
Gilbert  A.  Cullis,  of  the  "Imperial  College  of 
Science  and  Technology,"  and  Dr.  J.  W.  Evans,  of 
the  Imperial  Institute  (to  whom  I  am  indebted 
for  the  preface),  for  the  kind  encouragement  and 
advice  they  have  given  me  whilst  my  little  book 
"  was  in  the  making  ". 

MARGARET  REEKS. 


IMPERIAL  COLLEGE  OP  SCIENCE  AND  TECHNOLOGY, 
May,  1908. 


CONTENTS. 


PAGE 

PREFACE  BY  DR.  JOHN  W.  EVANS vii 


AUTHOR'S  PREFACE 


CHAP. 

I.  INTRODUCTORY 


II.  ISOMETRIC  SYSTEM  —  PROJECTION  OF  THE  AXES      ...         9 

III.  ISOMETRIC  SYSTEM  (cont.)—  OCTAHEDRON—  CUBE—  BHOMBDO- 

DECAHEDRON—  COMBINATIONS        .....        -        18 

IV.  ISOMETRIC    SYSTEM    (cont.)  —  TRAPEZOHEDRON  —  COMBINATION 

WITH  KHOMBDODECAHEDRON        ......        27 

V.  ISOMETRIC    SYSTEM    (cont.)—  TRISOCTAHEDRON  —  COMBINATION 
WITH  CUBE  —  TETRAHEXAHEDRON  —  COMBINATION  WITH  CUBE 

—  WITH  OCTAHEDRON    -        -        -        -.-,..-        -        -        33 

VI.  ISOMETRIC  SYSTEM  (cont.}  —  HEXOCTAHEDRON  40 

VII.  ISOMETRIC  SYSTEM  (cont.}  —  PYRITOHEDRONS  +  AND  --  COM- 

BINATIONS --  h  PYRITOHEDRON     OBLITERATED     BY    OCTA-       * 

HEDRON  --  PYRITOHEDRON    OBLITERATED    BY    OCTAHED- 

RON —  DIPLOID,  COMPLEMENTARY  FORM-—COMBINATION        -        45 

VIII.  ISOMETRIC  SYSTEM  (cont.}—  TETRAHEDRON  —  TRISTETRAHEDRON 

—  TETRAGONAL   TRISTETRAHEDRON—  HEXAKISTETRAHEDRON        53 

IX.  TETRAGONAL  SYSTEM  —  AXES  —  SIMPLE  FORMS  —  COMBINATIONS        58 

X.   HEXAGONAL     SYSTEM  —  CONSTRUCTION     FOR    AXES  —  SIMPLE 

FORMS  —  COMBINATIONS         ..-----        72 

XI.  HEXAGONAL    SYSTEM    (cont.}  —  KHOMBOHEDRON  —  SCALENO- 

HEDRON  —  TRAPEZOHEDRON  ..•••---        83 
xiii 


xiv  CONTENTS 

CHAP.  PAGE 

XII.  OBTHOBHOMBIC    SYSTEM— AXES— SIMPLE    FOBMS— COMBINA- 
TIONS     92 

XIII.  MONOCLINIC    SYSTEM  —  CONSTBUCTION    OF    AXES  —  SIMPLE 

FORMS — COMBINATIONS 107 

XIV.  TBICLINIC  SYSTEM — CONSTBUCTION  OP  AXES — SIMPLE  FOBMS 

— COMBINATIONS 120 

A  CHAPTEB  ON  TWINS.  TWINNED  CBYSTALS  IN  THE  ISOMETBIC — 
TETBAGONAL — HEXAGONAL — OBTHOBHOMBIC  —  MONOCLINIC 
— AND  TBICLINIC  SYSTEMS 130 


ILLUSTRATIONS. 

PLATE  PAOB 

I.  ISOMETRIC  SYSTEM 11 

Construction  of  Axes. 

II.  ISOMETRIC  SYSTEM 19 

Pig.  1.  Octahedron  {111}— Fig.  2.  Cube  {100}— Fig  3. 
Rhombdodecahedron  {101] — Fig.  4.  Cube  and  Octa- 
hedron in  Equilibrium — Fig.  5.  Cube  and  Octa- 
hedron, latter  predominant. 

III.  ISOMETRIC  SYSTEM 23 

Fig.  1.  Rhombdodecahedron  and  Cube — Fig.  2.  Rhomb- 
dodecahedron  and  Octahedron. 

IV.  ISOMETRIC  SYSTEM 25 

Fig.  1.  Cube  and  Rhombdodecahedron — Fig.  2.  Cube, 
Octahedron  and  Rhombdodecahedron. 

V.  ISOMETRIC  SYSTEM 29 

Fig.  1.  Trapezohedron  {211}— Fig.  2.  Auxiliary  Construc- 
tion— Fig.  3.  Trapezohedron  and  Rhombdodeca- 
hedron. 

VI.  ISOMETRIC  SYSTEM  -       --  -        .       -       .  35 

Fig.  1.  Trisoctahedron  {212}— Fig.  2.  Trisoctahedron  and 
Cube. 

VII.  ISOMETRIC  SYSTEM  -        -       -        -    '    -       -       -       «       -        37 
Fig.  1.  Tetrahexahedron  {201}— Fig.  2.  Tetrahexahedron 
and    Cube— Fig.    3.     Tetrahexahedron     {401}    and 
Octahedron. 

VIII.  ISOMETRIC  SYSTEM  -  -_       *'      -       -       .       41 

Hexoctahedron  {312}. 

IX.  ISOMETRIC  SYSTEM  -       -       .......     .       .       .       .        47 

Fig.  1.  +  Pyritohedron  {210}— Fig.  2.  -  Pyritohedron 
{201}. 


xvi  ILLUSTEATIONS 

PLATE  PAGB 

X.  ISOMETRIC  SYSTEM        ........        49 

Fig.  1.  +  Pyritohedron  and  Octahedron  in  equilibrium — 
Fig.  2.  -  Pyritohedron  and  Octahedron. 

XI.  ISOMETRIC  SYSTEM        ........51 

Fig.  1.  Diploid  {321}— Fig.  2.  Complementary  form  {312} 
— Fig.  3.  Combination  of  the  two. 

XII.  ISOMETRIC  SYSTEM 55 

Fig.  1.  Tetrahedron,  positive  {111] — Fig.  2.  Tetrahedron, 
negative  {111}— Fig.  3.  Tristetrahedron  {211}. 

XIII.  ISOMETRIC  SYSTEM 57 

Fig.  1.  Tetragonal  Tristetrahedron  {221}— Fig.  2.  Hexa- 
kistetrahedron  {312}. 

XIV.  TETRAGONAL  SYSTEM 59 

Fig.  1.  Method  of  obtaining  proportion  of  c  axis — Fig. 
2.  Method  of  finding  axes  proportional  to  given  axes 
by  "  parallelism  ". 

XV.  TETRAGONAL  SYSTEM 61 

Fig.  1.  Unit  Pyramid  (1st  order)  {111} ;  c  axis  =  1-75 — 
Fig.  2.  Unit  Pyramid  (1st  order)  {111} ;  c  axis  =  -4— 
Fig.  3.  Diametral  Pyramid  (2nd  order)  {101} ;  c  axis 
=  -67 — Fig.  4.  Ditetragonal  Pyramid  (Zirconoid) 
{313}— Fig.  5.  Pyramid  of  the  3rd  order  {313}— 
Fig.  6.  Unit  Prism  {110}  with  base  {001}— Fig.  7. 
Diametral  Prism  {100}  with  base  {001}— Fig.  8. 
Ditetragonal  Prism  {310}  with  base  {001}. 

XVI.  TETRAGONAL  SYSTEM 63 

Fig  1.  Zircon  c  axis  =  '64.  Prism  {110} ;  Pyramid  {111}— 
Fig.  2.  Zircon  c  axis  =  -64.  Prism  {110} ;  Pyramids 
{331},  {111}. 

XVII.  TETRAGONAL  SYSTEM     -        -        - 65 

Fig.  1.  Apophyllite  c  axis  =  1-25.  Prism  2nd  order  {100} ; 
Unit  Pyramid  {111}— Fig.  2.  Eutile  c  axis  =  -64. 
Prisms  1st  and  2nd  order  {110},  {100} ;  Pyramids  1st 
and  2nd  order  {111},  {101] — Fig  3.  Apophyllite  c  axis 
=  1-25.  Prism  2nd  order  {100} ;  Pyramid  1st  order 
{111};  Basal  Pinacoid  {001}— Fig.  4.  Sphenoid 
assumed  c  axis  =  1*5. 


ILLUSTRATIONS  xvii 

PLATE  PAGE 

XVIII.  TETRAGONAL  SYSTEM  •        -        -        •  9  '  -        -        •       -        69 
Zircon  c  axis  =  -64.    Prisms  of  1st  {110}  and  2nd  {100} 
orders — Unit    Pyramid    {111}    and    Ditetragonal 
Pyramid  {811}. 

XIX.  HEXAGONAL  SYSTEM  .---.•• 73 

Fig.  1.  Construction  of  axes — Fig.  2.  Profile  showing 
apparent  depression. 

XX.  HEXAGONAL  SYSTEM  -        -   0 75 

Fig.  1.  Pyramid  of  the  1st  order  {lOllJ — Fig.  2. 
Pyramid  of  the  2nd  order  {1122}— Fig.' 3.  Prism  of 
the  1st  order  {10lO}— Fig.  4.  Prism  of  the  2nd 
order  {1120}. 

XXI.  HEXAGONAL  SYSTEM 77 

Fig.  1.  Dihexagonal  Pyramid  {2133} — Fig.  2.  Pyramid 
of  the  3rd  order  {2133}. 

XXII.  HEXAGONAL  SYSTEM 79 

lodyrite  c  axis  =  -8 — Prism  of  the  2nd  order  {1120} ; 
Pyramid  1st  order  {4041} — Pyramid  1st  order 
{4045};  Base  {0001}. 

XXIII.  HEXAGONAL  SYSTEM 81 

Connelite  c  axis  =  1-16— Prisms  1st  order  {1110},  2nd 
{1120}— Pyramid  1st  order  {1111};  Dihexagonal 
Pyramid  (Berylloid),  Axial  Ratio  11  :  2  :  13  :  3. 

XXIV.  HEXAGONAL  SYSTEM 85 

Fig.  1.  Hematite  c  axis  =  1-36;  +  Khombohedron 
{1011}— Fig.  2.  Calcite  c  axis  =  -85 ;  -  Rhombo- 
hedron  {0111} — Fig.  3.  Calcite  +  Rhombohedron 
cut  by  Basal  Plane — Fig.  4.  +  Rhombohedron 
{lOll}  truncating  edges  of  -  Rhombohedron  {022"l}. 

XXV.  HEXAGONAL  SYSTEM  -        -    :    -       i       -       -       -       -        87 

Fig.  1.  Calcite  c  axis  =  -85.  Unit  Prism  {1110}  and- 
Rhombohedron  {0112}— Fig.  2.  Dioptase  c  axis  = 
•53.  Prism  2nd  order  {1120}  and  -  Rhombohedron 
{0221}. 

XXVI.  HEXAGONAL  SYSTEM 89 

Calcite  c  axis  =  -85 ;  +  Scalenohedron  {2131}. 
b 


xviii  ILLUSTBATIONS 

PLATE  PAGE 

XXVII.  HEXAGONAL  SYSTEM 91 

Quartz  c  axis  =  1-1 ;  Right-handed  Trapezohedron 
{2131}. 

XXVIII.  OETHOBHOMBIC  SYSTEM 93 

Fig.   1.    Unit   Prism  {110};  Basal   Pinacoid  {001}; 
assumed  Axial  Ratio  -8:1: 1-3 — Fig.  2.  Macro- 
pinacoid  {100};  Macroprism  {210};  Unit  Prism 
{110} ;  Brachyprism  {120}. 
• 

XXIX.  OBTHOBHOMBIC  SYSTEM 95 

Fig.  1.  Macropinacoid  {100} ;  Brachypinacoid  {010} ; 
Basal  Pinacoid  {001} — Fig.  2.  Unit  Pyramid 
{111}— Fig.  3.  Macrodome  {101} ;  Brachypina- 
coid {010}— Fig.  4.  Brachydome  {Oil};  Macro- 
pinacoid {100}. 

XXX.  OBTHOBHOMBIC  SYSTEM 97 

Fig.  1.  Barite.  Axial  Ratio  -8  :  1  :  1-3 ;  Unit  Prism 
{110}  ;  Macrodome  {102}— Fig.  2.  Barite.  Unit 
Prism  {110} ;  Macrodome  {102} ;  Brachydome 
{Oil};  Basal  Plane  {001}— Fig.  3.  Brookite. 
Axial  Ratio  -8  : 1 :  -9  ;  Unit  Prism  {110} ;  Brachy- 
pyramid  {122}— Fig.  4.  Arsenopyrite.  Axial 
Ratio  -67  :  1  :  1-18  ;  Unit  Prism  {110};  Brachy- 
dome {011} — Fig.  5.  Arsenopyrite.  Unit  Prism 
{110} ;  Brachydome  {014}. 

XXXI.  OBTHOBHOMBIC  SYSTEM 99 

Fig.  1.  Childrenite.  Axial  Ratio  -77  : 1  :  -53 ;  Macro- 
pinacoid {100} ;  Brachypinacoid  {010} ;  Pyramid 
{121}— Fig.  2.  Calamine.  Axial  Ratio  -78  :  1  : 
•48 ;  Prism  {110} ;  Pyramid  {121} ;  Brachypinacoid 
{010} ;  Brachydome  {031}  ;  Macrodome  {301}— 
Fig.  3.  Epsomite.  Axial  Ratio  -99 : 1  :  -57 ;  Prism 
{110} ;  Sphenoid  {111}. 

XXXII.  OBTHOBHOMBIC  SYSTEM 105 

Chrysolite.  Axial  Ratio  '46  : 1  :  -58  ;  Macropinacoid 
{100};  Macrodome  {101} ;  Brachypinacoid  {010} ; 
Brachydome  {021};  Unit  Prism  {110};  Unit 
Pyramid  {111} ;  Basal  Pinacoid  {001}. 


ILLUSTRATIONS  xix 

PLATE  PAGE 

XXXIII.  MONOCLINIC  SYSTEM        -        -     •  .  •     -        .       -  109 

Fig.  1.  Construction  of  Axes,  assumed  Axial  Ratio 
•65  :  1  :  -55— Fig.  2.  Profile  of  a  and  c  Axes- 
Fig.  3.  Unit  Prism  {110};  Base  {001}— Fig.  4. 
Orthopinacoid  {100} ;  Clinopinacoid  {010} ;  Basal 
Pinacoid  {001}. 

XXXIV.  MONOCLINIC  SYSTEM        -        -        -        -        -        -        -      111 

Fig.  1.  Pyramid  {111}— Fig.  2.  Pyramid  {111}— Fig  3. 
Combination  of  Pyramids — Fig.  4.  Gypsum. 
Axial  Ratio  -7  :  1  :  -4;  0  =  80°  42'.  Unit  Prism 
{110} ;  Unit  Pyramid  {111} ;  Clinopinacoid  {OlOj — 
Fig.  5.  Sphene.  Axial  Ratio  -75:1: -85;  0  =  60° 
17'.  Unit  Pyramid  {111} ;  Unit  Prism  {110}. 
XXXV.  MONOCLINIC  SYSTEM  - 113 

Fig.  1.  Clinodomes  {Oil}— Fig.  2.  Orthodomes  {101} 
— Fig.  3.  Amphibole.  Axial  Ratio  55  : 1 :  -29 ;  ft 
73°  58'.  Unit  Prism  {110};  Clinopinacoid  {010}; 
Clinodome  {Oil}— Fig.  4.  Epidote.  Axial  Ratio 
1-58  :  1  :  1-8 ;  0  64°  37'.  Orthopinacoid  {100} ; 
Orthodome  {110} ;  Basal  Pinacoid  {001} ;  Unit 
Pyramids  {111}  {111}  {111}  {111}. 

XXXVI.  MONOCLINIC  SYSTEM -        -      117 

Orthoclase.     Axial  Ratio  -66  :  1  :  -56 ;  Angle  0  =  63° 
57'.  Unit  Prism  {110};  Clinoprism  {130};  Unit 
Pyramid    {111};     Clinopinacoid    {010};     Basal 
Pinacoid  {001} ;  Orthodomes  {101}  {201}. 
XXXVII.  TEICLINIC  SYSTEM  -        -        ...        .        -        -      123 

Construction  of  Axes. 
XXXVIII.  TBICLINIC  SYSTEM •       -        -      125 

Fig.  1.  Pyramid  {111}— Fig.  2.  Pyramid  {111}— Fig. 
3.  Pyramid  {111}— Fig.  4.  Pyramid  {111}— Fig.  5. 
Prism  {110}— Fig.  6.  Prism  {110}— Fig.  7.  Pina- 
coid {100}— Fig.  8.  Pinacoid  {010] — Fig.  9.  Macro- 
dome  {101}— Fig.  10.  Macrodome  {101}— Fig.  11. 
Brachydome  {Ollj — Fig.  12.  Brachydome  {Oil}. 
XXXIX.  TRICLINIC  SYSTEM -  127 

Fig.  1.  Axinite.  Axial  Ratio  -49  :  1  :  -48  ;  a  82°  54' ;  0 
91°  52' ;  y  131°  32'.  Macropinacoid  {100} ;  Two 
Unit  Prisms  {110}  {110};  Macrodome  {201} ;  Two 
Unit  Pyramids  {111}  {111};  Brachyprism {010}— 
Figs.  2,  3,  4.  Construction  for  Axes. 


xx  ILLUSTRATIONS 

PLATE  PAGE 

XL.  ISOMETRIC  SYSTEM        -        -        -     .  -        -        -        -        -      131 
Fig.  1.  Spinel,  Twinned  Octahedron.      Fig.   2.  Profile 
— Fig.  3.  Shows  Positions  of  Twinning   Axis  and 
Plane. 

XLI.  TETRAGONAL  AND  ORTHORHOMBIC  SYSTEMS  -  -  -  135 
Tetragonal : — Fig.  1.  Cassiterite  c  axis  =  '67,  showing 
Twinning  Plane  parallel  to  Face  {101}— Fig.  2. 
Twinned  Crystal — Fig.  3.  Profile  Construction. 
Orthorhombic  : — Fig.  4.  Aragonite.  Axial  Ratio  -66  : 
1  :  -72.  Twinning  Plane  parallel  to  Prism  of  60°— 
Fig.  5.  Twinned  Crystal. 

XLII.  HEXAGONAL  SYSTEM 139 

Calcite  c  axis  =  -85 — Fig.  1.  +  Scalenohedron  {2131}, 
showing  Twinning  Plane  parallel  to  Face  of  - 
Rhombohedron  {0221] — Fig.  2.  Profile. 

XLIII.  HEXAGONAL  SYSTEM 143 

Calcite  c  axis  =  -85— Fig.  1.  +  Scalenohedron  {2131}, 
Twinning  Plane  -  Rhombohedron  {022~1}— Fig.  2. 
Profile  for  finding  c  axis  of  Twinned  Portion. 

XLIV.    MONOCLINIC   AND   TRICLINIC    SYSTEMS     -  -  -.'-.-         147 

Fig.  1.  Gypsum,  Axial  Ratio  -69  :  1  :  -4 ;  0  80°  42' ; 
showing  Twinning  Plane,  Orthopinacoid  {100}— 
Fig.  2.  Twinned  Crystal — Fig.  3.  Labradorite, 
Axial  Ratio  '63  :  1  :  -55 ;  a  93°  23' ;  ft  116°  29' ;  7  89° 
59' ;  showing  Twinning  Plane,  Pinacoid  {010} — 
Fig.  4.  Twinned  Crystal. 


OF   THE 

(   UNIVERSITY    J 


CHAPTEK  I. 

INTRODUCTORY. 

THE  sole  object  of  this  little  book  is  to  assist 
the  student  of  Mineralogy  in  making  drawings  of 
the  crystal  forms  and  combinations  with  which  he 
has  to  deal.  Nothing  beyond  this  is  aimed  at. 
No  idea  is  entertained  of  competing  with  the  many 
standard  works  on  crystallography,  to  which  the 
student  is  referred  for  information.  Merely  solu- 
tion of  graphical  problems  is  attempted  here,  with 
the  hope  that  the  attempt  will  not  be  entirely 
without  value. 

In  order  to  draw  crystal  forms  successfully  it  is 
absolutely  essential  to  have  some  slight  knowledge 
of  geometrical  projection,  to  know  how  to  make 
the  simple  plan  and  elevation  of  a  solid,  in  ortho- 
graphic projection. 

It  must  be  assumed  therefore  that  the  student 
has  such  knowledge,  also  that  he  has  thoroughly 
mastered — from  the  text-books  on  the  subject — 
the  several  characteristics  of  the  six  systems  under 
which  crystals  are  classed,  with  the  different 
methods  of  notation  most  in  use. 

Throughout  the  problems  here  given,  Miller's 
symbols  will  be  employed,  Bravais-Miller  in  the 
hexagonal  system.  The  examples  are  all  typical. 

The  subject  of  geometrical  projection  will  only 


2  HINTS  FOE  CEYSTAL  DEAWING 

be  followed  so  far  as  it  is  helpful  for  crystal  pro- 
jection. Trigonometry  will  hardly  appear  except 
in  the  guise  of  geometry. 

Of  the  many  projections  used  by  the  geo- 
metrician, that  known  as  clinographic  has  been 
mainly  selected  by  the  mineralogist  as  suitable  for 
crystal  projection.  Modern  text-books  almost  in- 
variably adopt  it  for  their  illustration — therefore 
clinographic  projection  will  be  chiefly  considered. 

Crystal  projection  is  very  dependent  on  parallel 
lines.  For  drawing  such  lines  set  squares  are  most 
convenient.  All  acquainted  with  mechanical  draw- 
ing know  the  value  of  the  60°  and  45°  squares,  yet 
at  the  risk  of  appearing  tedious  we  will  venture  a 
few  hints  on  their  correct  use. 

Fig.  1  will  illustrate  it.  A  and  B  represent  the 
position  of  the  right  and  left  hands  whilst  the 
squares  are  being  set  for  drawing  parallel  lines. 

The  fourth,  third,  middle  fingers  and  thumb  of 
the  left  hand  are  holding  the  60°  set  square  firmly 
pressed  against  the  paper  on  which  the  series  of 
parallel  lines  is  to  be  drawn.  The  first  finger  is 
merely  touching  the  45°  square  lightly,  acting  as  a 
guide,  so  that  the  square,  the  hypotenuse  of  which 
rests  against  the  hypotenuse  of  the  60°  square,  can 
slide  backwards  and  forwards  as  required. 

The  right  hand  effects  the  shifting  movements, 
holding  the  ruling  pen  ready  to  assume  quickly  the 
second  position  B',  whilst  the  line  is  being  drawn. 

The  first  finger  of  the  left  hand  is  pressed  firmly 
during  the  ruling,  releasing  the  pressure  again  to 
enable  the  45°  square  to  be  shifted  for  a  second 
line. 


INTRODUCTORY  3 

Should  it  be  desired  to  draw  lines  at  right 
angles  to  those  first  drawn,  the  45°  square  is  shifted 
backwards,  the  edge  at  right  angles  to  that  first 
used  coming  into  play. 

Clinographic  projection  has  probably  been  so 
generally  chosen  for  crystal  representation,  because 
it  in  some  measure  combines  the  elevation  and  plan 


FIG.  1. 


of  orthographic  projection,  thus  making  one  figure 
instead  of  two,  suffice  for  the  portrayal  of  one 
solid. 

It  has  been  assumed  that  the  student  is  fairly 
well  acquainted  with  orthographic  projection,  that 
he  knows  the  relations  of  the  vertical  and  horizon- 
tal planes  on  which  the  elevation  and  plan  are 
projected,  that  he  knows  the  relation  of  the  pro- 


4  HINTS  FOR  CRYSTAL  DRAWING 

jectors  to  each  of  these  planes  and  to  the  object 
they  are  projecting ;  he  knows  that  the  projectors 
in  the  case  of  the  plan  fall  vertically  on  every  point 
and  line  of  the  object,  as  it  were  transpierce  it,  and 
convey  the  impress  to  the  horizontal  plane  below ; 
he  knows  that  in  the  case  of  the  elevation,  the  pro- 
jectors transpierce  the  object  horizontally,  and  carry 
its  lines  and  points  to  a  vertical  plane  behind  it, 
which  plane  is  at  right  angles  to  the  horizontal 
projectors. 

Clinographic  projection  much  resembles  ortho- 
graphic, with  some  variation.  It  is  made  only  on 
a  vertical  plane,  which  plane,  it  is  convenient  to 
imagine  passes  through  the  centre  of  the  object  to 
be  projected. 

The  clinographic  projectors  are  neither  vertical 
nor  horizontal,  they  are  inclined ;  but  though  them- 
selves inclined,  they  are,  and  this  is  important,  all 
contained  in  vertical  planes,  which  planes  are  at 
right  angles  to  the  plane  of  projection. 

The  projectors  just  as  in  the  case  of  the  eleva- 
tion of  orthographic  projection,  transpierce  the 
object  and  carry  its  points  to  the  plane  of  projec- 
tion, but  since  the  plane  is  taken  through  the  centre 
of  the  object,  all  points  in  front  of  the  plane  will 
necessarily  be  carried  by  the  projectors  downwards 
and  backwards  to  it ;  all  points  behind  the  plane  will 
be  carried  to  \\>,  forwards  and  upwards. 

These  relations  of  projectors  to  planes  of  pro- 
jection both  in  orthographic  and  clinographic  pro- 
jection will  be  clear  from  Fig.  2,  which  is  a  profile 
or  side  view  of  the  planes  of  projection — A  ortho- 
graphic, B  clinographic.  The  figure  a,  b,  c,  d  is  the 


INTKODUCTOKY 


side  view  of  a  cube  which  is  being  projected  by  the 
projectors — shown  by  dashed  lines— to  the  respec- 
tive planes  of  projection,  as  seen  in  profile  re- 
presented by  lines. 


a 


B 


FIG.  2. 


It  should  be  apparent  from  these  figures  that 
in  clinographic  projection  the  top  face  of  the  cube 
will  be  visible,  in  orthographic  projection  it  will  be 
represented  as  a  line. 

The  student,  being  acquainted  with  orthographic 
projection,  knows  that  the  projectors  are  at  right 
angles  to  the  plane  of  projection,  because  the  eye 
is  assumed  to  be  opposite  each  point  projected  ;  in 
clinographic  projection,  the  eye  or  point  of  vision 
is  supposed  raised  above  each  point  a  certain 
number  of  degrees,  for  which  reason  the  projectors 
are  represented  by  inclined  lines. 

In  orthographic  projection  the  object  may  be 
inclined  to  either  plane ;  in  clinographic  it  is 
always  placed  with  one  axis  in  a  vertical  position 
and  is  then  rotated  round  this  axis  towards  the 
left,  through  some  selected  angle. 


6  HINTS  FOE  CEYSTAL  DEAWING 

The  result  of  this  rotation  is,  that  side  faces 
become  visible,  just  as,  on  account  of  the  elevation 
of  the  eye,  the  upper  faces  are  visible. 

Fig.  3.  A  is  an  orthographic  elevation  of  an 
octahedron.  B  is  a  clinographic  projection  of 
the  same  form.  In  both  it  has  been  rotated  in  the 
direction  of  the  movement  of  the  hands  of  a  clock, 


B 

FIG.  3. 

on  its  upright  axis.  In  the  orthographic  projection 
that  axis  has  been  inclined  a  certain  number  of  de- 
grees to  the  horizontal.  In  the  clinographic  pro- 
jection the  axis  is  vertical  but  the  point  of  vision 
has  been  elevated. 

The  figures  are  almost  identical ;  there  is  one 
difference :  in  the  clinographic  drawing  the  object 
has  its  actual  height,  in  the  orthographic  the  height 
is  slightly  foreshortened. 

In  clinographic  projection  there  are  two  most 
important  angles — the  angle  of  rotation  of  the  object 
towards  the  left  and  the  angle  of  the  elevation  of 
the  point  of  vision. 

Both  these  angles  are  optional — the  draughts- 
man will  vary  them  according  to  the  kind  of  view 
he  wishes  to  obtain — he  may  wish  to  show  more  of 


INTEODUCTOEY  7 

the  side  or  of  the  top  of  any  particular  crystal ;  he 
will  then  choose  wider  angles. 

Nevertheless,  though  this  is  justifiable — in  some 
cases  expedient — certain  angles  have  been  selected 
as  suitable  for  crystal  representation,  and  because 
of  their  suitability  and  almost  invariable  acceptance 
it  is  well  to  adhere  to  them. 

These  conventional  angles  are  for  the  rotation 
18°  26',  for  the  elevation  of  the  eye  9°  28'. 

As  they  are  of  such  importance  they  must  be 
constructed  with  most  careful  accuracy.  To  achieve 
this  accuracy  a  certain  trigonometrical  ratio  of  the 
angles  comes  most  opportunely  to  our  aid.  This 
ratio  is  the  tangent. 

The  tangent  of  the  angle  18°  26'  =  1  : 3. 

The  tangent  of  the  angle    9°  28'  =  1  : 6. 

By  which  is  simply  meant,  stated  in  language 
not  trigonometrical,  that,  if  a  perpendicular  be  drawn 
at  any  point  on  either  of  the  lines  enclosing  the 
angle,  and  be  produced  to  cut  the  other  enclosing 
line,  the  proportion  of  the  perpendicular  to  the  line 
on  which  it  is  raised,  measured  from  its  base  to  the 
apex  of  the  angle,  will  be  in  the  case  of  angle  18°  26' 
as  1  : 3  ;  in  the  case  of  angle  9°  28'  as  1  :  6. 

Knowledge  of  these  ratios  greatly  aids  in  the  con- 
struction of  the  angles. 

It  is  applied  thus :  a  line  is  drawn  at  some 
point ;  on  it  a  perpendicular  is  raised,  the  pro- 
portion of  the  perpendicular  to  the  line  is  made  as 
1  to  3  or  as  1  to  6,  the  top  of  the  perpendicular  is 
joined  to  the  end  of  the  line  to  form  the  angle. 
Figs.  4  and  5  should  demonstrate  this  sufficiently. 

Since  the  words  orthographic  and  clinographic 


8 


HINTS  FOE  CKYSTAL  DKAWING 


are  somewhat  lengthy  for  constant  use,  they  will 
often  be  represented  by  C  and  0.  Since  also  the 
terms  horizontal  and  vertical  will  frequently  recur, 
they  will  be  replaced  by  H  and  V. 


FIG.  4. 


The  student  is  reminded  that  with  reference  to 
projection,  lines  parallel  to  the  side  edges  of  the 
paper  are  called  vertical ;  those  parallel  to  the  upper 
and  lower  edges  are  called  horizontal. 


CHAPTEK  II. 

ISOMETRIC  SYSTEM— PROJECTION  OF  THE  AXES. 

PLATE  I. 

AXES  all  at  right  angles,  all  equal,  all  lettered,  a, 
av  #2,  as.     #1  and  #2  horizontal ;  az  vertical. 

The  first  step  towards  the  construction  of  any 
crystal  form  is  the  construction  of  the  axes. 

The  construction  of  the  isometric  axes  in  C  pro- 
jection shall  be  the  first  problem  (see  Plate  I.). 

In  C  projection,  the  vertical  axis  will  be  a 
vertical  line  of  its  actual  length.  We  can  at  once 
draw  it,  placing  it  towards  the  bottom  of  the  paper 
so  as  to  leave  plenty  of  space  above,  making  it  the 
desired  length  and  lettering  its  4-  and  -  ends,  +  &3, 
—  #3,  as  shown  in  the  figure. 

The  projection  of  the  two  horizontal  axes  is  less 
simple ;  they  will  not  appear  as  mere  horizontal 
lines ;  the  #1  will  be  inclined  downwards  towards 
the  left,  its  4-  end  will  appear  depressed. 

The  inclination  towards  the  left  is  dependent  on 
the  angle  of  rotation,  the  apparent  depression  on 
the  angle  of  elevation  of  the  eye.  We  have  to  find 
the  amounts  of  inclination  and  apparent  depression. 

To  do  this  we  call  to  our  aid  an  O  plan  of  the 
axes. 

This  we  construct  above  the  C  projection  of  the 


10  HINTS  FOE  CEYSTAL  DEAWING 

as  axis  already  drawn.  Choose  a  point  0  (see 
Plate  I.)  vertically  above  it.  This  point  is  the  0 
plan  of  the  as  axis,  which  in  accordance  with  the 
rules  of  0  projection  will  be  represented  simply  by 
a  point. 

Through  0  draw  a  vertical  and  also  the  line  OA  l 
making  with  the  vertical  the  angle  of  rotation  18° 
26'. 

This  angle  should  be  constructed  according  to 
the  directions  given  (see  Fig.  4). 

Make  the  line  OAl  equal  the  semi-axial  length. 

Through  0  draw  line  OA  2  at  right  angles  to 
OA19  making  it  equal  OAl  in  length. 

These  two  lines  OA  i  and  OA^  are  the  plans  of 
the  semi  a^  and  $2  axes  in  the  required  position  in 
orthographic  projection. 

We  want  these  axes  in  C  projection. 

We  already  have  the  vertical  az  axis  in  line 
4-  03,  -  as.  Bisect  this  axis  and  through  the  centre 
draw  the  H]im  X'Y'. 

Drop  vertical  projectors  from  the  +  ends  of  the 
H  axes  of  the  0  plan.  Somewhere  on  these  pro- 
jectors, below  the  level  of  Xf  Yf  will  lie  the  +  ends 
of  the  horizontal  axes  in  C  projection  ;  the  problem 
we  have  to  solve  is  to  find  their  exact  position  on 
these  projectors. 

The  simplest  method  is  as  follows  : — 

Through  point  0  of  the  0  plan  draw  the  H  line 
X  Y;  produce  the  vertical  through  Al  to  cut  this 
line  in  0',  divide  the  distance  GO  into  half  as 
shown,  then  by  means  of  the  dividers  convey  the  \ 
distance  to  the  C  projection,  marking  it  downwards 
from  the  line  X'  Yr  on  the  vertical  projector  dropped 


W 


ro 


v  Q 


+A 


-3 
PLATE  I. 

ISOMETRIC  SYSTEM. 
Construction  of  Axes. 


12  HINTS  FOR  CRYSTAL  DRAWING 

from  Al  in  point  R.  Point  R  will  be  the  position 
of  the  +  end  of  the  at  axis  in  C  projection. 

The  «!  axis  will  be  a  line  from  R  through  the 
centre  point  of  the  »3.  Produce  it  beyond  that 
axis  and  make  the  produced  portion  equal  the 
portion  on  the  near  side ;  this  will  complete  the  a^ 
axis. 

The  a2  axis  is  found  similarly.  Produce  the 
vertical  though  +  Az  of  the  0  plan  to  cut  the  line 
XYm  0".  Next  draw  through  +  A2  the  line  to  e 
at  right  angles  to  the  A2  axis  and  parallel  to  the  A^ 
axis.  Divide  the  distance  eO'  into  half  and  con- 
vey the  half  distance  to  the  C  projection,  marking 
it  downwards  from  X'Y  on  the  vertical  dropped 
from  +  A  2  in  R.^  R  is  the  position  of  the  4-  end  of 
the  Oz  axis  in  C  projection.  Join  R  to  centre  point 
of  the  vertical  axis  and  produce  the  line  an  equal 
length  beyond ;  this  line  is  the  #2  axis  in  C  pro- 
jection. 

This  completes  the  problem ;  we  have  the 
isometric  axes  in  C  projection. 

This  is  the  simplest  construction  for  the  C  pro- 
jection of  the  isometric  axes ;  it  may  be  easily 
mastered,  taken  on  trust,  used  mechanically.  With 
modification  it  may  be  applied  to  all  the  other  five 
systems,  so  that  without  much  trouble  and  with 
very  little  thought  many  of  the  simpler  forms  may 
be  drawn. 

Nevertheless  explanation  of  the  several  steps 
will  be  demanded. 

It  will  at  once  be  seen  how  the  0  plan  aids  the 
C  projection  much  in  the  same  way  as  it  aids  the 
orthographic  elevation. 


ISOMETBIC  SYSTEM  13 

From  the  0  plan  we  get  the  amount  of  fore- 
shortening consequent  on  the  rotation  to  the  left. 
The  vertical  projectors  dropped  from  the  +  ends  of 
the  Al  and  A2  axes  convey  this  to  the  C  projection 
below. 

But  the  0  plan  does  not  indicate  directly  the 
degree  of  foreshortening  of  the  axes  consequent  on 
the  elevation  of  the  point  of  sight. 

We  obtain  at  once  from  the  0  plan  how  much 
to  the  right  or  left  of  the  centre  the  +  ends  of  the  H 
axes  will  lie.  We  do  not  at  once  know  the  distance 
below  the  centre  they  will  appear  to  fall.  We  know 
they  will  lie  somewhere  on  the  vertical  projectors 
dropped  from  the  O  plan.  We  do  not  know  the 
exact  position  on  these  projectors. 

Let  us  first  consider  what  the  projection  of  the 
horizontal  axes,  the  a^  and  #2  would  be  supposing 
the  point  of  vision  had  not  been  raised  through 
9°  28'.  They  would  appear  in  C  projection  as  a 
horizontal  line,  passing  through  the  centre  of  the 
vertical  axis.  The  0X  axis  would  apparently  coin- 
cide with  the  a2  for  a  portion  of  the  length  of  the 
latter,  but  though  in  reality  of  the  same  length  it 
would  appear  shorter.  Its  4-  end  would  be  where 
the  V  projector  from  the  Al  axis  of  the  0  plan  cuts 
the  H  line  X'Y  of  the  C  projection.  The  +  end 
of  the  a.2  would  be  where  the  V  projector  from  A2 
cuts  the  same  line. 

But  the  elevation  of  the  point  of  vision  causes 
a  proportional  apparent  depression  of  the  +  ends  of 
the  two  horizontal  axes. 

Suppose  the  plane  of  projection  and  the  pro- 
jectors could  be  viewed  from  the  side  or  in  profile 


14  HINTS  FOE  CEYSTAL  DEAWING 

(Fig.  2,  Plate  I.  shows  such  a  view).  The  plane  of 
projection  is  represented  simply  by  the  vertical  line 
through  ORRf.  The  projectors  appear  as  inclined 
lines.  These  inclined  lines  make  the  angle  of 
elevation  9°  28'  with  any  horizontal  line  drawn  at 
right  angles  to  the  plane  of  projection. 

Draw  the  line  NO,  making  the  angle  9°  28'  with 
OP'  (for  construction  of  the  angle  see  Fig.  5) ;  all 
the  projectors  will  be  parallel  to  NO. 

Now  from  the  0  plan  we  take  the  distance  0', 
+  Alt  which  is  the  distance  the  +  end  of  the  Al 
axis  is  in  front  of  the  plane  of  projection.  We 
mark  this  distance  on  the  profile  in  OP' \  draw 
P'R'  parallel  to  NO.  K  gives  us  the  distance  the 
end  of  the  axis  will  appear  to  fall  below  the  level 
of  the  centre  0.  OR'  marked  on  the  vertical 
dropped  from  +  A1  of  the  0  plan  to  the  C  pro- 
jection will  give  the  exact  position  for  the  +  end 
of  the  #!  axis. 

The  +  end  of  the  a2  is  found  similarly.  The  dis- 
tance +  A  2  0"  is  marked  on  the  profile  in  OP ; 
the  projector  is  drawn  parallel  to  NO  to  cut  the 
plane  of  projection  in  R.  OR  is  the  amount  of 
apparent  depression  of  the  -f-  end  of  the  a2  axis  to 
be  marked  on  the  vertical  projector  dropped  from 
+  A  2  of  the  0  plan  in  R.  R  is  the  -f  end  of  the 
a,2  axis  in  C  projection. 

But  the  profile  is  not  necessary  for  construction 
and  all  unnecessary  work  is  best  dispensed  with. 
We  can  after  all  obtain  the  apparent  depression  from 
the  O  plan. 

For  as  a  consequence  of  the  tangential  relations 
of  the  angles  selected  for  rotation  and  elevation, 


ISOMETRIC  SYSTEM  15 

because  the  tangent  of  9°  28  is  ^,  whilst  that  of 
18°  26'  is  ^,  we  can  get  the  amount  of  depression  for 
any  point  in  a  very  simple  way :  we  have  only  to 
take  one-sixth  part  of  a  line  passing  through  the 
point  to  line  XY  &t  right  angles  to  that  line  ;  this 
sixth  part  will  be  the  amount  of  apparent  depression 
to  be  marked  downwards  on  the  vertical  projector 
below  the  line  X'Y'  of  the  C  projection. 

But  since  to  get  the  exact  sixth  we  must  either 
set  proportional  compasses  or  use  some  geometrical 
method,  and  since  such  compasses  may  not  be  at 
hand,  and  the  geometrical  method  or  the  more 
primitive  trials  by  the  dividers  method,  both  take 
time,  we  can  find  the  amount  in  the  case  of  the  + 
end  of  the  0,1  axis  by  dividing  the  distance  00'  into 
half  and  taking  the  half. 

Any  point  whose  apparent  depression  we  wish 
to  obtain  may  be  made  the  end  of  a  line  parallel  to 
the  Al  axis.  The  +  end  of  the  A2  is  the  end  of 
the  line  to  e,  which  is  parallel  to  the  A1  axis  ; 
J  eQP  will  give  the  amount  of  apparent  depression 
for  the  4-  end  of  the  a2  axis  in  C  projection.  It 
is  not  necessary  to  draw  the  line  to  et  we  need 
only  mark  the  point. 

An  easy  test  for  the  accuracy  of  the  construc- 
tion of  the  axes  is  to  draw  verticals  through  the 
ends  of  the  a±  and  #2  axes  and  horizontals  through 
the  ends  of  the  alt  The  space  between  the  H  lines 
taken  with  the  dividers  should  measure  twice  into 
the  space  between  the  verticals  through  the  ends 
of  the  di  axis  ;  and  the  space  between  the  verticals 
through  the  ends  of  the  a^  axis,  taken  with  the 
dividers,  should  measure  three  times  into  the 


16  HINTS  FOK  CKYSTAL  DBAWING 

space  between  the  verticals  through  the  ends  of 
the  #2  axis. 

It  will  be  noticed  that  only  half  the  0  plan  of 
the  axes  has  been  drawn,  for  since  half  supplies  all 
necessary  points,  no  more  is  needed,  the  fewer  lines 
used  for  construction  the  better ;  for  construction 
lines  are,  as  it  were,  the  scaffolding  necessary  whilst 
a  building  is  in  progress,  to  be  removed  when  it  is 
complete. 

In  all  figures  given  the  back  edges  will  be 
shown  dotted,  in  order  that  the  entire  form  or 
combination  of  forms  may  be  perfectly  realised. 
The  front  edges  will  be  shown  by  thick  lines,  the 
construction  lines  will  be  kept  fine. 

The  axes  will  always  be  indicated  by  chain 
lines  ;  when  the  figure  embraces  several  forms  the 
axial  ratio,  the  proportion  of  the  axes,  will  always 
be  shown  in  the  centre  of  the  figure. 

There  is  no  occasion  to  repeat  the  construction 
for  the  axes  each  time  ;  a  method,  which  we  will  call 
"  parallelism,"  may  be  used  for  reproducing  them. 

Let  the  axes  be  constructed  once  with  the 
utmost  care  and  accuracy  and  of  a  fair  length  on 
a  piece  of  thin  paper ;  the  vertical  axis  must  be 
absolutely  parallel  to  the  right-hand  edge. 

Draw  on  the  card  or  paper  on  which  the  pro- 
jection is  to  be  made  a  perfectly  vertical  line. 
Place  the  right-hand  edge  of  the  pattern  sheet 
accurately  against  this  line,  holding  it  firmly  in 
position,  whilst  a  set  square  is  adjusted*  in  turn  to 
each  of  the  horizontal  axes  and  lines  drawn  by  the 
parallel  method  from  a  point  selected  for  the  centre 
of  the  figure  to  be  projected.  These  lines  will  give 


ISOMETKIC  SYSTEM  17 

the  directions  of  the  horizontal  axes.    This  done  the 
pattern  sheet  can  be  removed  and  the  axes  completed. 

Should  it  be  desired  that  the  axes  of  the  draw- 
ing be  of  different  lengths  from  those  of  the  pattern 
this  may  be  attained  by  a  further  application  of 
"  parallelism". 

Before  removing  the  pattern  sheet  adjust  a  set 
square  to  touch  the  apex  of  the  pattern  vertical 
axis  and  the  4-  end  of  either  of  the  H  axes,  then 
having  marked  the  desired  length  of  the  new  verti- 
cal axis,  shift  the  square  without  changing  its  in- 
clination, by  keeping  it  well  against  the  edge  of 
the  guiding  square  until  it  touches  the  apex  of  the 
new  vertical  axis ;  it  will  cut  off  the  correct  pro- 
portional length  on  the  new  H  axis.  The  length 
of  the  other  H  axis  may  be  obtained  in  a  similar 
way ;  reference  to  Fig.  1  should  make  this  method 
clear  (see  also  Plate  XIV.). 

This  method  if  carried  out  with  great  care  is 
possibly  more  exact  than  that  of  pricking  through, 
because  the  latter,  by  constant  use,  wears  the 
paper,  causing  inaccuracy. 

Several  methods  may  be  employed  for  the  solu- 
tion of  the  problems  offered  by  crystal  drawing. 

These  methods  we  will  call  :— 

The  parallel  method  or  parallelism. 

The  method  of  symmetry  or  repetition. 

The  method  of  plane  intersection. 

The  profile  method. 

The  word  " repetition"  has  been  substituted  for 
symmetry  to  obviate  confusion  between  geometrical 
and  crystallographic  symmetry  and  will  be  used 
throughout  the  explanations. 


CHAPTEK  III 

ISOMETRIC  SYSTEM  (CONT.)— OCTAHEDRON— CUBE— RHOMBDO- 
DECAHEDRON— COMBINATIONS. 

PLATES  II.  TO  IV. 

BEFORE  attempting  to  project  any  crystal  it  is 
absolutely  necessary  to  realise  it,  that  is  to  say, 
to  have  a  clear  idea  of  its  proportions  and  their 
interrelation  before  the  mental  vision. 

To  attain  such  an  idea  it  is  useful  to  have  a 
model,  but  if  such  a  thing  is  not  at  hand  the 
practised  draughtsman  can  construct  a  mental  image 
from  certain  angles  and  ratios  found  in  the  text- 
books ;  to  do  this  is  an  exercise  for  the  reasoning 
and  imaginative  faculties.  The  power  to  realise  a 
form  not  seen,  is  a  power  well  worth  acquiring. 

To  get  such  a  clear  vision  of  a  crystal  form,  it 
is  seldom  necessary  to  have  exact  measurements  of 
every  edge  and  angle.  The  repetition  of  symmetry 
in  most  of  the  systems  obviates  this. 

In  the  isometric  system  very  few  data  will 
generally  be  required,  practice  alone  will  enable 
the  student  to  select  such  as  will  best  aid- him  to 
realise  a  form  or  combination  of  forms.  • 

It  will  be  assumed  that  the  student  knows  the 
several  forms  by  name  or  possesses  a  text-book  of 
mineralogy  in  which  he  may  find  them  described. 

Though  the  cube  is  figured  first  in  the  books,  it 

18 


PLATE  II. 

ISOMETRIC  SYSTEM. 
FIG.  1.  Octahedron  {111}. 
FIG.  2.  Cube  {100}. 
FIG.  3.  Rhombdodecahedron  {101}. 
FIG.  4.  Cube  and  Octahedron  in  Equilibrium. 
FIG.  5.  Cube  and  Octahedron,  latter  predominant. 
2* 


20  HINTS  FOK  CKYSTAL  DKAWING 

will  be  advisable  to  take  the  octahedron  on  account 
of  its  simplicity,  its  symbol  {111}  shows  that  all  its 
faces  cut  all  three  axes  at  unity. 

When  once  the  construction  for  the  axes  has 
been  mastered,  drawing  the  octahedron  will  offer  no 
further  difficulty. 

Plate  II.  Fig.  1  shows  the  octahedron  formed 
by  joining  the  ends  of  the  axes. 

The  cube  is  drawn  by  the  first  of  the  methods 
mentioned  on  page  17,  "parallelism"  (see  Plate 
II.  Fig  2). 

Construct  the  axes  or  transfer  them,  by  the 
method  given  on  page  16. 

Through  the  ends  of  the  a^  axis  draw  lines 
parallel  to  the  az  axis,  cut  them  by  lines  parallel  to 
the  a},  drawn  through  the  ends  of  the  a*  Through 
the  points  of  cutting  raise  verticals,  make  the 
verticals  equal  the  true  axial  length,  complete  the 
cube  by  joining  the  ends  of  the  verticals. 

Make  the  near  edges  thick  lines,  the  back  ones 
dotted. 

If  the  drawing  is  made  with  due  accuracy  the 
side  face  of  the  cube  will  appear  ^  the  width  of 
the  front  face.  The  upper  face  will  appear  J  the 
width  of  the  side  face  and  £  the  width  of  the  front 
face. 

Plate  II.  Fig.  3  is  the  rhombdodecahedron. 

Having  obtained  the  axes,  join  the  ends  of  the 
ax  and  a*  Join  them  also  to  the  ends  of  the  az  axis  ; 
in  other  words,  complete  the  octahedron  lightly, 
divide  each  edge  into  exactly  half  in  the  points 
e,f,  (/,  h,  i  ;  through  these  points  draw  lines  parallel 
respectively  to  the  three  axes,  i.e.,  lines  parallel  to 


ISOMETKIC  SYSTEM  21 

the  a.2  through  e,f,g\  parallel  to  the  a^  through  h  ; 
to  the  #3  through  i,  k. 

Where  these  lines  intersect  will  be  the  solid 
angles  of  the  form. 

The  figure  is  completed  by  the  second  method 
(see  p.  17)  by  "  repetition ".  The  draughtsman 
knowing  the  points  which  are  vertically  below  or 
on  the  same  level  with  those  found,  that  is  to  say, 
knowing  thoroughly  the  symmetry  of  the  rhombdo- 
decahedron,  completes  it  by  marking  off  similar 
lengths  on  selected  parallel  lines.  He  will  test  the 
accuracy  of  his  work  by  trying  with  the  set  squares 
whether  the  several  points  lie  on  lines  parallel  to 
the  axes  ;  he  will  not  need  actually  to  draw  the  lines, 
as  he  becomes  skilful  he  will  more  and  more  dis- 
pense with  auxiliary  lines,  merely  marking  the  re- 
quired points  of  intersection  by  a  sharply  defined 
pencil  dot. 

Fig.  4  shows  a  combination  of  forms — the  cube 
and  the  octahedron ;  they  are  in  so-called  equili- 
brium, the  faces  of  the  octahedron  meet  at  the 
middle  points  of  the  edges  of  the  cube. 

The  construction  is  at  once  apparent.  Lines 
are  drawn  through  the  ends  of  each  axis  parallel 
to  the  other  two  axes  ;  they  intersect  in  points 
e>  f>  9i  h,  i,  k,  /,  m,  etc.  ;  join  these  points  for  the 
edges  of  the  combination. 

Fig.  5  shows  the  same  combination,  but  in  this 
case  the  octahedral-  faces  greatly  predominate. 
The  cube  cuts  the  a^  axis  in  point  p.  Draw  a  short 
vertical  through  p  to  cut  the  two  edges  of  the 
octahedron  at  points  1  and  2.  From  these  points 
the  intersections  of  the  two  forms  can  be  drawn ; 


22  HINTS  FOK  CKYSTAL  DEAWING 

they  are  each  parallel  to  one  edge  of  the  octa- 
hedron. 

For  the  upper  cubic  face  point  3  can  be  found 
by  taking  the  distance  from  the  end  of  the  a^  axis  to 
point  1  and  marking  it  from  the  vertex  of  the  az  axis 
in  point  3.  The  edges  of  intersection  can  again 
be  drawn  from  3  parallel  to  octahedral  edges. 

All  the  other  cubic  faces  may  be  found  similarly. 

Plate  III.  Fig.  1.  The  rhombdodecahedron 
cut  by  cubic  faces  gives  an  extreme  example  of  the 
use  of  "  parallelism  ". 

Draw  first  the  rhombdodecahedron  quite  lightly. 

It  is  assumed  that  the  cubic  faces  cut  the  axes 
in  points  #,  /,  g.  These  points  can  be  marked  from 
the  ends  of  the  respective  axes.  A  line  through 
point  e  parallel  to  the  a2  axis,  and  a  line  through 
point  /  parallel  to  the  vertical  axis  will  cut  lines 
drawn  through  the  centres  of  the  rhombdodeca- 
hedral  faces  and  by  cutting  them  will  give  points 
through  which  the  intersection  edges  of  the  two 
forms  can  be  drawn  parallel  to  the  several  axes  ; 
careful  attention  to  the  figure  will  make  this  clear. 

Many  more  lines  than  are  necessary  have  been 
shown  in  this  example  to  illustrate  how  "  parallel- 
ism "  may  be  used  to  test  accuracy,  but  to  the  ex- 
perienced worker  such  a  network  of  lines  is  of 
course  needless ;  by  mere  adjustment  of  the  set 
squares  he  will  check  the  parallelism. 

Fig.  2  shows  the  rhombdodecahedron  in  com- 
bination with  the  octahedron.  The  rhombdo- 
decahedral  faces  truncate  the  octahedral.  Point 
p  is  where  the  apex  of  the  octahedron  would  be  if 
it  had  not  given  place  to  the  other  form. 


(1) 


(2) 


PLATE  III. 

ISOMETRIC  SYSTEM. 

FIG.  1.  Rhombdodecahedron  and  Cube. 
FIG.  2.  Rhombdodecahedron  and  Octahedron. 
a  {100}  o{lll} 


24  HINTS  FOE  CKYSTAL  DKAWING 

Having  lightly  drawn  the  rhombdodecahedron 
join  the  +  ends  of  the  0,1  and  a2  axes,  bisect  the 
joining  line  and  join  the  point  of  bisection  to  the  + 
end  of  the  a3  axis.  From  point  p  draw  a  line  parallel 
to  that  from  apex  of  vertical  axis  to  bisection  of  line 
from  +  end  of  di  to  +  end  of  #2  axis.  Where  the 
line  from  point  p  cuts  a  short  vertical  through  the 
point  of  bisection  of  line  ah  az,  will  give  a  point  e, 
through  which  passes  an  edge  of  union  of  an  octa- 
hedral and  rhombdodecahedral  face. 

It  will  readily  be  seen  how  the  combination  can 
be  completed  by  lines  parallel  to  octahedral  edges. 
E/hombdodecahedral  edges  form  all  the  solid  angles. 

Plate  IV.  Fig.  1  shows  the  cube  in  combination 
with  the  rhombdodecahedron,  the  former  being 
greatly  predominant. 

Having  drawn  one  quadrant  of  the  cube,  quite 
lightly  as  shown-,  assume  that  the  rhombdodecahe- 
dron cuts  the  upper  cubic  face  in  point  e.  One 
edge  of  intersection  of  the  two  forms  will  of  course 
pass  through  e,  and  since  all  the  intersection  edges 
will  be  parallel  to  cubic  edges  we  can  at  once  draw 
the  edge  m,  n  through  e. 

Through  e  draw  line  XXf  parallel  to  x,  x'  an  octa- 
hedral edge ;  through  x  the  -f  end  of  the  #2  axis 
raise  a  vertical  to  cut  x,  x'  in  /;  /  is  another  point 
through  which  passes  an  intersection  edge. 

By  parallelism  find  point  Z* ',  the  point  on  the 
#!  axis  correspondent  to  point  x  on  the  a2.  Draw 
Z'Z  parallel  to  x,  y,  an  octahedral  edge.  Z'Z  will 
meet  the  upper  cubic  face  in  Z  and  the  front  one 
in  t.  Through  Z  and  t,  intersection  edges  can  be 
drawn. 


PLATE  IV. 

ISOMETRIC  SYSTEM. 

FIG.  1.  Cube  and  Rhombdodecahedron. 
FIG.  2.  Cube,  Octahedron  and  Rhombdodecahedron. 
a  {100}  o  {111}  d  {101}. 


OF   THE 

UNIVERSITY 


26  HINTS  FOE  CKYSTAL  DKAWING 

The  edges  through  e  and  Z  meet  in  point  n. 

Bisect  the  portion  of  XX'  between  e  and /and 
the  portion  of  ZZ'  between  Z  and  t ;  through  the 
bisections  draw  lines  parallel  to  the  intersection 
edges.  The  meeting  of  these  lines  gives  a  point 
P.  P  will  be  a  solid  angle  of  the  rhombdodecahed- 
ron. 

Draw  the  short  lines  nn't  nn",  parallel  to 
octahedral  edges.  Join  P  to  n,  n',  n"  for  rhomb  - 
dodecahedral  edges.  Complete  the  figure  by  "  par- 
allelism" and  "repetition". 

The  small  figure  below,  a  combination  of  cube, 
octahedron  and  rhombdodecahedron,  need  not  be 
described.  The  student  will  doubtless  be  able  to 
construct  such  a  combination. 


CHAPTEE  IV. 

ISOMETRIC  SYSTEM  (CONT.)— TRAPEZOHEDRON— COMBINA- 
TION WITH  RHOMBDODECAHEDRON. 

PLATE  V. 

THE  problem  of  the  trapezohedron  depends  on 
plane  intersection  for  solution. 

In  the  method  of  plane  intersection,  the  axial 
planes  play  a  most  important  part;  the  student 
should  therefore  endeavour  to  master  and  clearly 
realise  their  positions  and  interrelation. 

In  the  isometric  system  there  are  three  such 
planes.  They  are  all  at  right  angles — two  are 
vertical,  one  is  horizontal.  They  intersect  in 
lines  coincident  with  the  axes. 

The  first  vertical  plane  passes  through  the  a2 
and  #3  axes. 

The  second  vertical  plane  is  at  right  angles  to 
the  first,  it  passes  through  the  a±  and  az  axes ;  its 
intersection  with  the  first  plane  is  a  line  coincident 
with  the  03  axis. 

The  third  plane  is  horizontal,  it  is  at  right 
angles  to  both  the  others,  it  passes  through  the  a^ 
and  a.2  axes  ;  its  intersections  with  the  other  two 
planes  are  lines  coincident  with  the  a:  and  a2  axes. 

The  intersections  of  these  planes  with  each 
other  and  with  all  other  planes  will  be  represented 
by  lines. 

27 


28  HINTS  FOE  CEYSTAL  DEAWING 

The  student  will  recognise  that  these  three 
planes  divide  the  crystal  form  into  octants,  but 
they  are  not  limited  by  the  edges  of  that  form, 
they  extend  beyond  it  on  all  sides,  indefinitely. 

By  considering  the  relations  of  the  several 
crystal  faces  to  these  planes  we  are  able  to  under- 
stand and  project  the  form. 

The  trapezohedron  we  will  take  for  illustration 
has  the  symbol  {211}  (see  Plate  V.  Fig.  1). 

We  must  first  imagine  the  trapezohedron 
divided  into  eight  portions  by  the  axial  planes. 
These  portions  correspond  in  position  to  the  eight 
faces  of  the  octohedron.  We  will  consider  the 
construction  of  the  upper  right-hand  octant  only ; 
having  solved  this  problem  the  construction  of  the 
others  will  be  matter  of  "parallelism"  and  "re- 
petition ". 

It  will  be  seen  that  each  octant  is  formed  by 
three  faces.  These  faces,  bounded  by  edges,  are 
limited  portions  of  three  planes  which  must  be 
imagined  to  extend  beyond  the  faces. 

These  planes  are  neither  vertical  nor  horizontal, 
they  are  oblique. 

If  the  student  has  a  model  of  a  trapezohedron 
and  will  hold  it  so  that  one  axis  is  vertical,  another 
horizontal  and  slightly  turned  towards  the  left 
(reference  to  a  figure  of  a  trapezohedron  in  any 
text-book  will  enable  him  to  get  the  position),  he 
will  see  the  three  oblique  faces  which  compose  the 
upper  right-hand  octant,  and  will  by  a  stretch  of 
the  imagination  be  able  to  realise  them  as  portions 
of  oblique  planes  which  extend  beyond  their  limits. 

Oblique  plane  No.  I.  passes  through  the  +  end 


ISOMETRIC  SYSTEM. 
FIG.  1.  Trapezohedron  {211}. 
FIG.  2.  Auxiliary  Construction. 
FIG.  3.  Trapezohedron  and  Rhombdodecahedron. 
n  {211}  d  {101}. 


30  HINTS  FOE  CKYSTAL  DRAWING 

of  the  tfa  axis,  slants  towards  the  +  ends  of  the 
«2  and  az  axes,  but  at  such  an  inclination  that  it 
never  meets  them  ;  if,  however,  they  were  produced 
to  double  their  semi-length  it  would  meet  them  and 
at  the  same  time  would  meet  the  vertical  axial  plane 
which  contains  the  &3  and  a2  axes.  The  face  having 
the  symbol  {211}  is  the  limited  portion  of  this 
plane. 

Oblique  plane  No.  II.  passes  through  the  +  end 
of  the  a.2  axis,  slants  towards  the  +  ends  of  the 
di  and  #3  axes,  would  never  meet  either  but  would 
meet  both  produced  to  double  their  semi-length, 
meeting  at  the  same  time  the  vertical  axial  plane 
that  bisects  the  trapezohedron  through  the  ax  axis. 
The  face  having  the  symbol  {121}  is  the  limited 
portion  of  this  plane. 

Oblique  plane  No  III.  passes  through  the  +  end 
of  the  $3  axis,  slants  downwards  towards  the  other 
two  axes,  at  such  an  inclination  that  it  would  never 
meet  either,  but  would  meet  both  produced  to  double 
their  semi-length  and  would  likewise  meet  the 
horizontal  axial  plane  which  divides  the  trape- 
zohedron into  the  upper  and  lower  half.  The  face 
having  the  symbol  {112}  is  the  limited  portion  of 
this  plane. 

Each  of  these  three  oblique  planes  has  an  inter- 
section, or  as  it  is  called,  a  trace,  on  the  three  axial 
planes  which  pass  through  the  centre  of  the  form. 

If  we  can  find  two  such  traces  for  each  plane 
we  can  define  the  octant. 

How  shall  we  find  these  traces  ? 

From  the  +  end  of  the  a^  axis  draw  a  line  to 
cut  the  +  end  of  the  a2  axis  produced  to  twice  its 


ISOMETKIC  SYSTEM  31 

semi-length.     This  line  is  the  horizontal  trace  of 
the  first  of  the  oblique  planes. 

Find  next  the  H  trace  of  the  second  oblique 
plane.  It  starts  from  the  4-  end  of  the  a2  axis  to 
cut  the  +  end  of  the  a^  axis  produced  to  twice  its 
semi-length.  Now  these  two  H  traces  meet  in  a 
point  2,  the  portions  between  the  point  of  meeting 
2  and  the  +  ends  of  the  a^  and  a2  axes  are  the  two 
horizontal  edges  of  the  octant. 

To  find  the  two  front  edges  we  need  the  traces 
of  the  first  of  the  oblique  planes  and  of  the  third  of 
these  planes  on  the  V  plane  which  passes  through 
the  #!  axis.  These  traces  are  found  by  a  quite 
similar  method. 

Join  the  4-  end  of  the  ai  axis  to  a  point  on  the 
#3  axis  produced  to  twice  its  semi-length.  Join 
the  +  end  of  the  az  axis  to  the  +  end  of  the  a^ 
produced  to  twice  its  semi-length.  These  lines 
give  the  desired  traces,  they  meet  in  a  point  1. 
The  portions  between  the  point  of  meeting  and 
the  ends  of  the  axes  are  the  front  edges  of  the 
octant. 

The  side  edges  are  found  in  the  same  manner. 
By  joining  the  +  ends  of  the  a.2  and  az  axes  to  their 
semi-lengths  produced,  the  edges  are  the  portions 
between  the  point  of  intersection  3  of  the  traces 
and  the  +  ends  of  the  axes. 

The  octant  at  present  appears  as  Fig.  2.  From 
point  1,  draw  a  line  to  the  -h  end  of  the  a2  axis  pro- 
duced to  double  its  semi-length.  From  point  2 
draw  a  line  to  the  -I-  end  of  the  az  axis  produced 
to  double  its  semi-length.  These  lines  will  inter- 
sect in  a  point  P  which  will  be  the  solid  angle  of 


32  HINTS  FOE  CKYSTAL  DRAWING 

the  octant.  The  point  must  be  joined  to  point  3 
for  the  intersection  of  the  upper  and  right-hand 
lower  faces  {112}  and  {121}. 

The  right-hand  upper  octant  is  now  complete. 
The  others  may  be  constructed  by  "  parallelism  "  and 
"  repetition  ". 

The  lines  in  the  figure  passing  upwards,  are 
shown  merely  to  indicate  how  all  the  intersections 
of  the  planes  which  meet  the  a±  and  a.2  axes  at 
unity  would  meet  the  #8  at  twice  its  semi-length. 
They  give  a  good  test  for  accuracy. 

The  figure  below,  Fig.  3,  shows  a  combination  of 
trapezohedron  with  rhombdodecahedron. 

Assume  point  P  to  be  the  apex  of  the  rhomb- 
dodecahedron. 

Having  constructed  the  trapezohedron  by  rules 
given,  find  the  points  where  the  rhombdodecahedral 
faces  d  cut  the  edges  of  the  trapezohedron.  This  is 
done  by  drawing  the  lines  PP',  PP" ,  and  P'P" 
parallel  to  octahedral  edges  ;  they  cut  trapezohedral 
edges  in  points  1,  2,  3,  4,  5,  6.  Next  draw  lines 
from  the  angles  of  the  trapezohedron  t,  t',  t"  to  the 
centre  ;  the  intersections  of  these  with  the  lines  PP' 
PP' ',  P'P"  give  points  through  which  lines  parallel 
to  the  different  axes  may  be  drawn,  which  will  cut 
the  edges  of  the  trapezohedron,  giving  the  points 
7,  8,  9,  10, 11,  12 ;  lines  connecting  these  with  1,  2,  3? 
4,  5,  will  give  the  intersections  of  the  two  forms  and 
complete  the  construction  of  one  octant,  the  others 
may  be  found  by  the  usual  methods. 


CHAPTEK  V. 

ISOMETRIC  SYSTEM  (owr.)— TRISOCTAHEDRON— COMBINA- 
TION WITH  CUBE— TETRAHEXAHEDRON— COMBINATION 
WITH  CUBE— WITH  OCTAHEDRON. 

PLATES  VI.,  VII. 

THE  trisoctahedron  offers  little  difficulty  of  con- 
struction (Plate  VI.  Fig.  1). 

Begin  by  constructing  an  octahedron  every  edge 
of  which  will  appear  in  the  trisoctahedron. 

The  completion  of  the  latter  form  presents  one 
problem  for  solution ;  how  to  find  the  point  where 
the  three  oblique  planes  of  each  octant  meet ;  the 
solid  angle  X. 

Plane  intersection  will  be  the  method  to  employ. 

We  have  to  find  the  intersections  of  the  three 
faces  with  the  symbols  {212},  {122}  and  {221}. 

As  in  the  trapezohedron  we  have  to  find  the  H 
traces  of  the  planes  {212}  {122};  but  since  it  is 
sometimes  inconvenient  to  extend  the  axes  to  double 
their  length  we  may  use  a  proportional  method  and 
instead  of  producing  them  divide  their  semi-length 
into  half. 

Having  carefully  so  divided  the  semi  at  and  #2 
axes  join  the  half  points  of  each  to  the  end  of  the 
other.  Now  these  joining  lines  will  show  the  direc- 
tion of  the  H  traces  of  the  planes ;  we  need  the  H 
traces  themselves ;  they  are  lines  drawn  through 

33  3 


34  HINTS  FOE  CKYSTAL  DKAWING 

the  end  of  the  a^  axis  parallel  to  that  which  joins 
its  half  to  the  end  of  the  az  axis,  and  through  the 
end  of  the  a2  axis  parallel  to  that  which  joins  its 
half  to  the  end  of  the  a^  These  last  drawn  lines 
intersect  in  point  P.  A  line  from  the  intersections 
of  the  traces,  point  P,  to  the  +  end  of  the  az  axis 
gives  the  intersection  of  the  planes.  It  remains  to 
find  point  X,  the  point  where  the  plane  {221}  meets 
and  obliterates  the  other  two  planes. 

Form  a  parallelogram  inside  the  octahedron 
edges  by  drawing  through  the  ^  points  of  the  axes 
lines  parallel  to  those  edges.  Bisect  the  lines  of 
this  parallelogram  and  join  the  bisections  to  the 
+  end  of  the  az  axis.  A  line  drawn  from  the 
centre  of  the  octahedral  edge  g  parallel  to  that  to 
the  end  of  the  as  axis  will  cut  the  intersection  of 
the  planes  {212}  {122}  in  point  X.  Join  X  to  the 
+  ends  of  the  three  axes  and  the  octant  will  be 
complete. 

The  solid  angles  of  the  other  seven  octants  may 
be  found  by  "  parallelism  "  ;  by  lines  parallel  to  the 
respective  axes  cutting  lines  drawn  from  the  centres 
of  the  octahedral  edges  parallel  to  those  from  the 
inner  parallelogram  to  the  -h  end  of  the  as  axis. 

Plate  VI.  Fig.  2.  The  trisoctahedron  cut  by 
cube. 

First  find  the  trisoctahedron.  It  will  of  course 
not  be  necessary  to  complete  all  the  lines. 

Assume  the  cube  faces  to  cut  the  axes  of  the 
trisoctahedron  in  points  P,  Pf,  P"' 

Through  these  points  draw  lines  parallel  to  the 
«i,  az,  and  az  axes  to  cut  the  octahedral  edges  at 
1,  2,  3,  4 ;  1',  2',  3',  V ;  V ',  2",  3",  4". 


c\j 


I 

1 
I 

I 

8 


el 

H  ? 


36  HINTS  FOE  CRYSTAL  DRAWING 

To  find  points  where  the  cube  faces  cut  the  other 
intersections  of  the  trisoctahedral  faces  join  point  1 
to  3  and  4,  bisect  the  joining  lines,  draw  lines 
through  the  bisection  points  and  P,  taking  the  lines 
on  to  cut  the  trisoctahedral  edges  in  points  5  and 
6.  Find  7  and  8  similarly.  Complete  the  cubic 
face  by  joining  the  8  points.  Finish  the  figure  by 
"  parallelism  "  and  "  repetition  ". 

The  tetrahexahedron  (Plate  VII.  Fig.  1)  is  easy 
of  construction.  The  form  having  the^symbol  {201} 
has  been  selected  for  illustration. 

Let  lines  be  drawn  from  the  +  end  of  the  a^ 
axis  to  cut  each  of  the  other  axes  at  twice  their 
semi-lengths  produced,  and  lines  from  the  +  ends 
of  the  a2  and  a3  axes  to  cut  the  di  axis  produced 
to  twice  its  semi -length.  These  lines  will  intersect 
in  points  pp',  through  which  draw  lines  parallel  to 
the  axes  as  shown  in  the  figure.  These  lines  will 
be  edges  of  the  crystal  faces,  where  they  intersect 
each  other  will  be  solid  angles.  All  that  remains 
to  do  is  to  join  these  solid  angles  and  complete 
the  solid  by  " parallelism "  and  "repetition". 

The  combination  with  the  cube  is  constructed 
thus :  Assume  the  upper  face  of  the  obliterating 
cube  to  cut  off  the  solid  angle  of  the  tetrahexahedron 
at  the  level  of  point  X ;  through  this  point  draw 
lines  parallel  respectively  to  the  a^  and  a.2  axes 
where  these  lines  cut  those  used  for  the  construc- 
tion of  the  tetrahexahedron,  which  pass  from  the 
apex  of  the  vertical  axis  to  twice  the  semi-length 
of  the  0j  and  a2  axes  produced ;  will  give  two 
points  1  and  2  through  which  the  intersection 
edges  of  cube  and  tetrahexahedron  pass. 


(3) 

PLATE  VII. 
ISOMETRIC  SYSTEM. 

Fid.  1.  Tefcrahexahedron  {201}. 

FIG.  2.  Tetrahexahedron  and  Cube. 

FIG.  3.  Tetrahexahedron  {401}  and  Octahedron. 


38  HINTS  FOR  CRYSTAL  DRAWING 

Other  two  points,  3  and  4,  may  be  found  by 
short  lines  drawn  through  1  and  2,  parallel  to  an 
octahedral  edge  to  cut  the  lines  mounting  from  the 
+  ends  of  the  at  and  a2  axes. 

Careful  thought  will  sufficiently  direct  the  com- 
pletion of  the  figure  by  "  parallelism,"  "  repetition  " 
and  edge  intersection. 

Fig.  3.  The  combination  with  the  octahedron  in 
which  the  latter  prevails  is  found  thus  :— 

Draw  the  octahedron.  Mark  a  point  V  on  the 
vertical  axis  and  assume  it  to  be  the  apex  of  the 
tetrahexahedron. 

Find  the  corresponding  points  v',  v"  ;  t,  if  on  the 
other  axes. 

The  symbol  of  the  tetrahexahedron  is  in  this  case 
{401}-  We  have  to  find  the  point  x,  where  a  face 
of  the  tetrahexahedron  meets  the  front  octahedral 
edge.  This  is  attained  by  drawing  a  line  from  V 
to  R7,  W  being  the  distance  of  t  from  0  the  centre, 
multiplied  four  times.  Line  V  W  is  the  line  of 
inclination  of  the  face  {104}  and  x  the  point  where 
that  face  cuts  the  octahedral  edge. 

The  rest  of  the  problem  is  a  matter  of  "  parallel- 
ism "  and  "  repetition  ". 

The  distance  tfx  marked  up  the  front  octahedral 
edge  from  $f,  gives  point  x1 ',  the  correspondent 
point  oc"  on  the  back  octahedral  edge  may  be 
found  in  a  similar  way. 

of  can  be  transferred  to  z  and  zf  by  "  parallel- 


ism". 


u,  <  ?/',  may  all  be  found  by  "  parallelism  "  and 
measuring  off  corresponding  equal  distances. 

Having  thus  determined  all  points  on  octahedral 


ISOMETKIC  SYSTEM  39 

edges  where  the  two  forms  meet,  we  have  next  to 
find  intermediate  points  such  as  r. 

Through  points  #,  u,  z,  etc.,  draw  lines  parallel 
to  the  several  axes  so  as  to  form  the  lightly  drawn 
squares.  Join  the  points  t,  t/  v,  tfy  etc.,  to  the 
corners  of  the  squares.  These  diagonals  of  the 
squares  will  be  the  intersections  of  the  different 
tetrahexahedral  faces  ;  we  have  to  cut  off  these 
intersections  where  they  meet  the  octahedral  faces. 

For  this  purpose  draw  through  t  line  t,  s  parallel 
to  the  &2  axis,  t,  s  is  the  line  of  intersection  of  the 
face  {401}  of  the  tetrahexahedron  with  the  hori- 
zontal axial  plane,  s  is  the  point  where  this  inter- 
section meets  the  octahedral  edge,  s  then  is  a 
point  which  lies  in  both  the  face  {401}  and  the 
octahedral  face  ;  it  is  common  to  both,  therefore 
line  od  s  is  the  intersection  of  these  faces,  but  at 
T  this  intersection  meets  the  face  {410};  we  have 
therefore  to  join  r  to  u. 

The  similar  points  /,  r",  rf",  may  now  be  easily 
obtained  by  "  parallelism  "  and  "  repetition". 


CHAPTEK  VI. 

ISOMETRIC  SYSTEM  (CONT.)— HEXOCTAHEDRON. 

PLATE  VIII. 

THE  hexoctahedron  is  a  somewhat  complicated  form. 
The  student  must  before  undertaking  its  projec- 
tion carefully  master  its  proportions  and  realise  the 
relation  of  the  several  faces  to  those  of  the  octahed- 
ron. 

The  form  having  the  symbol  {312}  has  been 
chosen  for  illustration.  (Plate  VIII.) 

It  will  be  well  to  make  this  figure  an  example 
of  how  when  a  crystal  is  very  intricate,  much  of  the 
work  of  construction  may  be  done  outside  the  actual 
drawing  by  working  from  the  0  plan. 

Having  found  the  axes  or  transferred  them,  we 
can  at  once  construct  the  near  and  distant  central 
edges  LMN,  Lmn,  using  proportional  points  to  avoid 
extension  of  the  axes.  They  will  evidently  be  lines 
from  the  +  and  =  ends  of  the  a\  and  az  axes 
parallel  to  lines  drawn  from  the  ^  to  the  J  points 
of  these  axes  as  the  several  face  symbols  direct. 

We  have  next  to  find  the  horizontal  edges. 
These  as  we  know  will  be  the  horizontal  traces  of 
the  planes  to  which  they  belong. 

Since  these  traces  cut  very  obliquely  in  the  C 
projection,  their  exact  intersection  being  on  that 

40 


^s.. 

OF    THE  \ 

UNIVERSITY    J 

OF  / 


PLATE  VIII. 

ISOMETRIC  SYSTEM. 
Hexoctahedron  {312}. 


42  HINTS  FOE  CEYSTAL  DEAWING 

account,  difficult  to  determine,  we  may  avail  our- 
selves of  the  0  plan. 

Instead  of  drawing  these  traces  directly  on  the 
C  projection  in  the  usual  manner  by  lines  from  the 
ends  of  the  axes  to  cut  the  other  axes  produced  at 
the  proper  ratio  for  the  respective  planes,  we  may 
draw  them  on  the  0  plan,  from  the  +  ends  of  the  #1 
and  a2  axes  parallel  to  lines  drawn  from  propor- 
tional points. 

Then  from  the  intersections  on  the  0  plan,  which 
will  be  more  accurate  because  more  acute,  we  can 
drop  them  by  projectors  to  cut  lines  bisecting  the 
angles  between  the  H  axes  in  the  C  projection. 

To  complete  the  upper  right-hand  octant  we  have 
to  find  the  two  edges  from  the  -f  ends  of  the  a2  and 
#3  axes ;  this  is  easily  done  by  the  proportional 
points.  The  central  point  X  where  all  the  planes 
forming  the  octant  meet  offers  more  difficulty. 

To  find  it  we  will  employ  what  we  have  termed 
the  " profile  method". 

Suppose  it  possible  to  cut  through  the  crystal  by 
a  clean  vertical  cut,  through  the  solid  angle  of  the 
octant  and  then  to  take  a  side  or  profile  view  of 
the  section  thus  revealed. 

Such  a  view  would  be  of  immense  advantage, 
because  from  it  we  could  at  once  measure  the  dis- 
tance the  point  X, — the  point  we  require, — the  point 
of  the  solid  angle — is  from  the  vertical  axis  and  also 
the  height  it  is  above  the  horizontal  plane  con- 
taining the  &J  and  a2  axes. 

Such  a  view  we  can  construct  reasoning  as 
follows :  Our  section  to  obtain  the  profile  passes 
through  line  PQ  in  the  C  projection ;  through  PQ[ 


ISOMETBIC  SYSTEM  43 

of  the  0  plan.  We  need  only  concern  ourselves  with 
that  portion  lying  above  PQ  in  the  C  projection,  the 
lower  portion  can  always  be  found  by  "  repetition  ". 

We  take  the  line  P'Q  of  the  0  plan ;  it  is  one 
true  line  for  the  profile.  We  have  another  line  PL 
the  semi  as  axis  ;  that  too  is  a  true  line  for  the  profile. 
One  other  fact  we  have  towards  its  construction : 
these  two  lines  are  at  right  angles  to  one  another. 
The  next  step  therefore  is  to  draw  P'L'  equal  PL 
at  right  angles  to  PQf  on  the  O  plan. 

Two  more  lines  we  require,  L'X'  and  X'Q ; 
they  also  must  be  of  the  true  length  in  the  profile. 
How  shall  we  ascertain  their  length  ?  If  we  have  a 
model  which  is  dependable  enough  we  can  measure 
them.  The  safer  means  of  arriving  at  their  length 
is  by  construction. 

We  know  that  the  planes  which  have  XQ  for 
their  intersection  both  meet  the  as  axis  at  3  times  its 
own  semi-length  from  the  centre,  therefore  we  can 
find  the  inclination  of  the  line  of  intersection.  If 
on  the  0  plan  we  draw  a  line  from  ($  to  P'L'  pro- 
duced to  3  times  its  length  we  shall  have  the  desired 
inclination. 

To  prevent  making  the  drawing  extend  beyond 
the  limits  of  our  paper  we,  instead  of  producing  line 
P'L',  draw  the  line  of  inclination  from  Z  to  L',  Z 
being  the  proportional  point  for  Q  found  by  drawing 
lines  through  the  \  and  ^  points  of  the  a  and  a2  axes. 

A  line  parallel  to  ZL'  through  Ql  is  the  line  we 
want  for  our  profile. 

But  we  need  its  true  length.  How  shall  we 
arrive  at  it  ?  By  finding  the  true  inclination  of  the 
intersection  of  the  planes  {213}  and  {123}  and  draw- 


44  HINTS  FOE  CRYSTAL  DRAWING 

ing  a  line  from  Lf  at  this  inclination,  to  cut  off  the 
line  through  $'. 

To  find  the  true  inclination  of  this  intersection, 
first  find  point  Zf  on  the  0  plan.  Z'  is  the  propor- 
tional point  on  line  PfQf  through  which  the  H 
traces  of  planes  {213}  and  {123}  would  pass.  Z"is 
the  proportional  point  of  the  r/3  axis  of  the  profile 
through  which  they  would  both  pass.  Line  Z"Z ', 
gives  the  inclination  of  their  intersection.  A  line 
through  Lr,  the  apex  of  the  az  axis  of  the  profile 
parallel  to  Z"Z',  will  cut  off  the  line  from  Q  in  the 
point  X'. 

From  X'  draw  X'K'  at  right  angles  to  line  P'Q. 
Kf  will  be  the  point  in  the  0  plan  vertically  below 
X,  the  solid  angle  we  require.  Drop  K'  by  a  verti- 
cal projector  to  the  C  projection  to  cut  the  line  PQ 
in  K ;  then  from  K  mark  on  the  vertical  projector 
a  height  KX  =  K'X',  which  is  the  true  height  of 
the  solid  angle  X  above  the  horizontal  plane  through 
the  H  axes.  Join  X  to  the  other  points  of  the 
octant  already  found.  The  octant  is  then  complete. 

Complete  the  other  octants  by  "parallelism" 
and  "  repetition  ". 

The  angle  of  the  left-hand  back  octant,  for 
example,  is  found  by  making  Pk  =  PK  and  from 
k  raising  a  vertical  =  KX. 

The  student  will  perhaps  realise  the  profile  con- 
struction better,  if  he  imagines  a  clean  cut  made 
through  the  points  LXQ  and  then  imagines  it 
possible  to  turn  the  cut  open  on  line  PQ  acting  as 
a  hinge.  This  has  been  done  on  the  0  plan.  P'Q 
is  the  hinge  line,  the  face  of  the  clean  cut,  the  pro- 
file, is  bounded  by  lines  P'L'X'ty. 


CHAPTEK  VII. 

ISOMETRIC  SYSTEM  (CONT.)  — PYBITOHEDRONS  +  AND COM- 
BINATIONS— +  PYRITOHEDRON  OBLITERATED  BY  OCTA- 
HEDRON  PYRITOHEDRON  OBLITERATED  BY  OCTAHED- 
RON—DIPLOID,  COMPLEMENTARY .  FORM— COMBINATION. 

PLATES  IX.  TO  XI. 

THE  construction  of  the  +  pyritohedron  is  quite  easy 
(Plate  IX.  Fig.  1).  It  is  worked  by  plane  inter- 
section. The  form  with  the  symbol  {210}  has  been 
chosen  for  demonstration.  In  this  case  the  actual 
lengths,  not  proportional  ones,  have  been  used  for 
the  plane  intersection. 

A  word  of  direction  may  be  needed  as  to  the 
manner  of  finding  line  X  Y  the  intersection  of  planes 
{102}  and  {210}. 

Raise  a  vertical  00'  at  0,  the  point  of  inter- 
section of  the  line  drawn  from  the  +  end  of  the 
#!  axis  to  cut  the  a2  produced.  This  vertical,  shown 
at  OO',  is  the  intersection  of  plane  {210}  with  the 
vertical  axial  plane  through  the  as  and  a2  axes. 
Produce  the  edge  of  the  pyritohedron  LP  to  cut 
the  vertical  at  0' ;  then  a  line  drawn  from  (/  to 
X  will  be  the  intersection  of  the  planes  {210}  and 
{102K  because  X  and  Of  lie  in  both  those  planes. 

To  find  the  point  Y  and  the  intersection  PY 
produce  the  edge  through  the  4-  end  of  the  a*  axis  to 
meet  line  VZ  drawn  parallel  to  the  #2  axis  through 
the  tfj  axis  produced  to  twice  its  semi-length. 

45 


46  HINTS  FOK  CEYSTAL  DKAWING 

Join  Z,  which  is  a  point  common  to  both  the 
planes  {021}  and  {102},  to  P.  The  line  joining  will 
cut  X Of  in  Y ;  this  is  the  required  point. 

Complete  the  figure  by  "  parallelism  "  and  "  re- 
petition ". 

The  minus  pyritohedron  is  constructed  similarly 
by  plane  intersection  (Plate  IX.  Fig.  2). 

Point  0  is  found  by  producing  the  edge  through 
the  4-  end  of  the  a^  axis  to  cut  a  line  drawn  parallel 
to  the  di  through  the  a2  produced  to  twice  its  semi- 
length.  Join  the  point  of  intersection  /  thus  found 
to  J.  Then  through  the  +  end  of  the  az  axis  pro- 
duced ta,  twice  its  semi-length  draw  a  line  parallel 
to  the  #2  tovmeet  the  produced  vertical  edge  through 
the  +  end  of  the  a2  axis  in  L.  Join  the  point  of 
intersection  to  K ;  the  line  to  K  will  cut  the  line 
from  J  to  /  in  0,  the  point  required. 

The  student  should  carefully  realise  each  step  of 
the  plane  intersection  employed  for  these  two  figures. 

The  last  figure  is  not  quite  satisfactory.  Two 
faces,  the  left-hand  top  face  {012}  and  the  right- 
hand  lower  face  {012}  appear  as  single  lines.  It  is 
a  case  when  an  angle  other  than  the  conventional 
one,  might  be  preferred  for  the  elevation  of  the  point 
of  vision. 

Plate  X.  Fig.  1,  shows  the  +  pyritohedron  {210} 
combined  with  an  octahedron.  The  two  forms  are 
in  what  is  termed  equilibrium,  that  is,  are  equally 
present.  It  will  be  seen  at  once  how  the  edges  of 
intersection  are  found,  by  joining  those  angles  of 
the  pyritohedron  which  meet  in  any  of  the  axial 
planes.  Angles  not  meeting  in  these  planes  are 
obliterated  by  the  octahedral  faces. 


(1) 


(2) 

PLATE  IX. 

ISOMETRIC  SYSTEM. 
FIG.  1.    +  Pyritohedron  {210}. 
FIG.  2.    -  Pyritohedron  {201}. 


48  HINTS  FOE  CKYSTAL  DEAWING 

Fig.  2  is  the  —  pyritohedron  in  combination  with 
the  octahedron. 

We  found  in  constructing  Fig.  2,  Plate  IX.,  that 
the  usual  angle  for  the  elevation  of  the  eye  gave  an 
unsatisfactory  view  of  the  -  pyritohedron.  It  will 
be  well  to  substitute  a  wider  angle ;  15°  shall  be 
selected. 

With  this  angle  we  cannot  find  the  proportionate 
apparent  depression  by  taking  half  distances ;  the 
profile  of  the  angle  must  be  constructed  and  the 
amounts  of  apparent  depression  taken  from  it  as 
shown. 

The  complete  axial  construction  is  represented. 

We  will  assume  the  left-hand  angle  of  the 
octahedral  face  {111}  to  be  at  C. 

Draw  the  octahedral  edge  CC'.  It  cuts  line 
k,  I,  the  inclination  line  in  the  horizontal  axial 
plane,  of  the  face  {120}  of  the  pyritohedron  in  point 
p,  and  cuts  the  edge  of  the  pyritohedron  passing 
through  the  +  end  of  the  #1  axis  at  p'. 

The  line  pp'  is  the  portion  of  the  octahedral 
edge  intercepted  between  the  pyritohedral  faces 
{201}  and  {120}. 

Draw  the  octahedral  edge  G  V.  Line  C'  V  cuts 
the  inclination  line  of  the  face  {201}  in  point  n  and 
the  edge  of  the  pyritohedron  passing  through 
the  +  end  of  the  az  axis  in  nf ;  the  portion  of  the 
octahedral  edge  between  these  points  is  the  portion 
required. 

By  joining  (7  to  V  we  get  point  v'.  v'v"  is  a  portion 
of  another  octahedral  edge  present  in  the  combina- 
tion. Joining  these  points  pp' ,  nn',  rV,  we  have 
an  octahedral  face.  The  others  can  be  obtained  by 


(2) 
PLATE  X. 

ISOMETRIC  SYSTEM. 

FIG.  1.    +  Pyritohedron  and  Octahedron  in  equilibrium. 
FIG.  2.    -  Pyritohedron  and  Octahedron. 


50  HINTS  FOE  CEYSTAL  DEAWING 

similar  reasoning  and  by  "parallelism"  and  "re- 
petition ". 

Plate  XL  Fig.  1,  shows  the  construction  for  the 
diploid  by  plane  intersection.  The  form  having  the 
symbol  {321}  has  been  selected.  The  intersections 
of  planes  {213}  {213}  and  {132}  with  the  three  axial 
planes  are  shown  complete. 

Lines  XX'  ^  XX" ,  X'X" ,  are  the  intersections  or 
traces  of  the  plane  containing  the  face  {321}  with  the 
three  axial  planes.  XX ',  X'X" ,  are  the  vertical 
traces.  XX"  is  the  horizontal  trace. 

Lines  ZZ ',  ZZ' ',  Z'Z"  are  in  like  manner  the 
traces  of  the  plane  containing  the  face  {213}- 

Lines  FF',  FF",  F'F",  are  the  traces  of  the 
plane  containing  the  face  {132}- 

•  The  student  will  have  little  difficulty  in  compre- 
hending the  relative  positions  of  these  planes  from 
their  intersections  with  the  axial  planes.  He  will 
see  at  once  the  method  of  finding  the  intersecting 
edges,  which  is  shown  in  the  case  of  the  edge  e,  f 
by  the  line  e,  g,  drawn  from  e,  the  point  of  inter- 
section of  the  two  vertical  traces,  on  the  axial  plane 
through  the  a^  and  #3  axes,  to  g  the  corresponding 
intersection  of  -  the  vertical  traces  on  the  axial 
plane  through  the  axes  a2  and  #3.  e,  g  being  points 
common  to  either  plane,  line  e,  g  must  be  the  inter- 
section of  the  planes  and  of  the  faces  which  are 
portions  of  the  planes. 

Fig.  2.  The  complementary  form  is  constructed 
in  exactly  the  same  manner  by  finding  the  traces  of 
the  several  planes  involved,  on  the  axial  planes,  by 
reference  to  the  respective  face  symbols. 

Fig.  3.     The  combination  of  the  two  forms  is  a 


52  HINTS  FOE  CKYSTAL  DEAWING 

little  more  intricate.  It  is  worked  in  the  same 
way. 

First  the  solid  angles  which  are  numbered  may 
be  found ;  these  correspond,  as  indicated  by  the 
numbers,  to  the  solid  angles  of  the  original  forms, 
taken  alternately.  The  positions  of  the  edges  start- 
ing from  these  angles  will  remain  unchanged.  The 
edges  meeting  in  the  axial  planes  will  change  their 
direction. 

They  are  to  be  got  by  finding  the  points  of  inter- 
section of  the  traces  of  the  planes  in  which  they  lie. 

The  construction  for  edge  h,  i  is  given  in  dotted 
lines  as  an  example.  Proportional  points  have  been 
employed  to  save  space ;  it  h  is  parallel  to  i'hf,  the 
proportional  intersection  of  the  two  planes  contain- 
ing faces  {321}  and  {312}- 


CHAPTEE  VIII. 

ISOMETKIC  SYSTEM  (owr.)—  TETRAHEDRON— TRISTETRA- 

HEDRON— TETRAGONAL  TRISTETRAHEDRON— 

HEXAKISTETRAHEDRON. 

PLATES  XII.,  XIII. 

THE  tetrahedrons,  positive  and  negative,  are  very 
easy  of  construction  (see  Plate  XII.  Figs.  1  and  2). 

The  edges  are  drawn  through  the  ends  of  the 
axes  parallel  to  the  edges  of  the  octahedron. 

For  the  positive  tetrahedron  {111}  the  edge 
through  the  4-  end  of  the  az  axis  is  parallel  to  the 
octahedral  edge  from  the  +  end  of  the  &i  to  the  + 
end  of  the  a2  axis. 

For  the  negative  tetrahedron  {111}  the  edge 
through  the  —  end  of  the  az  axis  is  parallel  to  that 
octahedral  edge. 

The  positive  tristetrahedron  {211}  may  be  con- 
structed by  plane  intersection  from  the  positive 
tetrahedron. 

From  each  angle  draw  lines  to  intersect  the  axes 
produced  to  twice  their  semi-length ;  where  these 
lines  cut  each  other  will  be  the  solid  angles  of  the 
figure. 

It  is  not  necessary  to  draw  lines  to  the  ends  of 
all  three  axes  for  each  point ;  two  such  lines  will 
suffice  but  the  third  ensures  accuracy. 

Plate  XIII.  Fig.  1  gives  the  tetragonal  tris- 
tetrahedron {221}  (deltoid  dodecahedron). 

53 


54  HINTS  FOE  CRYSTAL  DRAWING 

Draw  lightly  the  upper  half  of  the  octahedron. 

The  next  step  is  to  find  the  points  X,  Xf,  X", 
X'". 

To  find  point  X.  It  will  be  situated  somewhere 
on  the  normal  to  the  octahedral  face  {111}-  The 
normal  will  be  a  line  passing  from  the  centre  of  the 
axes  0,  through  the  central  point  of  the  face  {111}. 
The  centre  of  the  face  is  the  intersection  of  two  lines 
joining  the  middle  points  of  any  two  edges  of  the 
face  to  the  opposite  angles. 

Next  draw  the  H  trace  of  the  plane  {122}  from 
the  -h  end  of  the  #2  axis  to  the  +  end  of  the  a^  pro- 
duced to  twice  its  semi-length  A,  t. 

To  cut  this  trace  draw  linp  Oc.     Join  c  to  the 
+  end  of  the  a3  axis ;  the  joining  line  will  be  the 
intersection  of  planes  {212}  {122} ;  the  point  where, 
it  cuts  the  normal  will  be  the  solid  angle  X. 

Draw  line  xxr  parallel  to  Oc  ;  mark  point  of 
as  far  to  the  left  of  the  V  axis  as  x  is  to  the  right. 

Mark  point  d  on  the  V  axis  as  far  below  the 
centre  as  e  is  above.  Through  ef  draw  two  lines 
x'' ',  x'tf  and  /,  g,  parallel  respectively  to  xx'  and  to 
the  octahedral  edge  st.  g  is  a  point  vertically 
under  x  and  by  drawing  gx"  parallel  to  the  a2  axis 
we  get  point  #",  and  marking  the  distance  ex"  to 
the  right  of  the  V  axis  we  get  xm. 

We  now  have  the  four  solid  angles  x,  of,  x",  x"f. 
Draw  from  these  several  points,  lines  through  the 
central  points  of  octahedral  edges  ;  where  these 
lines  cut  will  be  the  remaining  angles  required  k,  i,  I, 
etc.  i  has  however  been  found  preferably,  by  draw- 
ing a  line  from  k,  parallel  to  st,  to  cut  the  line  from 
x ;  the  line  from  x  cuts  too  obliquely. 


xn. 


ISOMETRIC  SYSTEM. 
FIG.  1.  Tetrahedron,  positive  {111}. 
FIG.  2.  Tetrahedron,  negative  {111}. 
FIG.  3.  Tristetrahedron  {211}. 


56  HINTS  FOE  CEYSTAL  DEAWING 

The  construction  of  the  hexakistetrahedron,  Fig. 
2,  is  soon  described.  The  form  selected  is  {312}. 
Find  first  all  the  edges  of  the  hexoctahedron,  that 
pass  through  the  axes  and  the  points  X,  X,  X,  X, 
the  solid  angles  of  alternate  octants  as  directed  in  the 
construction  for  the  hexoctahedron  (see  Chap.  VI. 
Plate  VIII.).  Join  points  X  to  the  points  of  inter- 
section of  the  edges.  From  all  the  X  points,  produce 
alternate  edges  to  meet  alternate  edges  from  the 
other  X  points,  as  shown  in  the  figure  and  thus  the 
form  is  completed. 

Exercises — Problems  for  solution. 
Isometric  System.1     Garnet  {110}  {211}. 

Tetrahexahedrons  {320}  {410}  {530}. 

Tetragonal  tristetrahedron  {443}. 

Trapezohedron  {311}. 

1  All  these  combinations  are  found  in  the  collection  of  100 
crystal  models  of  rock-forming  minerals  arranged  by  Dr.  K. 
Busz  and  sold  by  Dr.  F.  Krantz  of  Bonn. 


(2) 

PLATE  XIH. 

ISOMETRIC  SYSTEM. 

FIG.  1.  Tetragonal  Tristetrahedron  {221}. 
Fia.  2.  Hexakistetrahedron  {312}. 


CHAPTEK  IX. 

TETRAGONAL  SYSTEM— AXES— SIMPLE  FORMS— COMBINA- 
TIONS. 

PLATES  XIV.  TO  XVIII. 

THE  axes  of  the  tetragonal  system  are  all  at  right 
angles.  The  horizontal  axes  are  equal  and  are 
lettered  as  in  the  isometric  system,  alt  az.  The 
vertical  axis  varies  in  length  and  is  lettered  c. 

Since  the  #x  and  a2  axes  in  the  tetragonal  system 
are  at  the  same  angle  and  in  the  same  ratio  as  in 
the  isometric,  that  is,  are  at  right  angles  and  equal 
to  one  another,  it  is  evident  that  no  different  con- 
struction for  finding  them  is  needed. 

The  c  axis,  however,  which  takes  the  place  of 
the  &3  axis  of  the  isometric  system,  is  of  a  different 
length,  longer  or  shorter  than  the  other  two. 

How  to  get  the  relative  proportion  of  the  c  axis 
is  easily  seen  from  Plate  XIV.  Fig.  1. 

Suppose  the  three  axes  constructed  according  to 
the  directions  given  for  the  isometric  system.  As- 
sume the  c  axis  required  to  have  the  proportion  *6  ; 
all  we  have  to  do  is  to  divide  the  semi  a3  axis  of  the 
isometric  system  into  decimal  parts,  *6  will  be  the 
+  end  of  the  c  axis  which  is  marked  off  below  the 
centre  for  the  -  c. 

Fig.  2,  Plate  XIV.  shows  a  convenient  method 
by  which  having  once  constructed  the  axes  we  may 

58 


PLATE  XIV. 

TETRAGONAL  SYSTEM. 

FIG.  1.  Method  of  obtaining  proportion  of  c  axis. 
FIG.  2.  Method  of  finding  axes  proportional  to  given  axes  by 
"  parallelism  ". 


60  HINTS  FOE  CKYSTAL  DBA  WING 

draw  sets  of  axes  either  longer  or  shorter,  having 
the  c  axis  in  the  same  ratio,  or  in  a  different  ratio. 

Let  lines  0  alf  a2,  #3  be  the  semi-axes  of  the 
isometric  system.  Divide  a3  into  decimal  parts. 

Suppose  we  wish  to  draw  tetragonal  axes  of  ^ 
the  length,  the  ratio  of  the  c  axes  being  1  :  *6. 

Draw  a  line  a'2  in  the  position  desired  parallel  to 
the  a2  axis  but  J  the  length,  and  a  line  parallel  to 
the  at  axis  through  the  centre. 

Now  arrange  the  set  squares  as  shown  by  the 
solid  lines ;  the  edge  of  the  45°  passing  exactly 
through  the  +  end  of  the  a2  axis  and  the  point  '6  of 
the  as  axis.  Slide  the  set  square  against  the  edge 
of  the  60°,  being  most  careful  that  the  latter  does 
not  shift. 

When  the  edge  of  the  45°  set  square  reaches 
the  +  end  of  the  a'2  axis,  the  new  a2  axis,  the  posi- 
tion shown  by  the  dotted  lines,  it  will  cut  the  new 
vertical  axis  in  point  c  which  will  be  in  the  correct 
ratio. 

The  new  Oi  axis  may  be  cut  off  the  correct  length 
by  arranging  the  edge  of  the  45°  set  square  to  pass 
through  the  +  end  of  the  original  a:  axis  and  the  + 
end  of  the  original  az  axis  and  then  sliding  it  as  be- 
fore on  the  60°  set  square  until  it  passes  through 
the  +  end  of  the  a'2  axis ;  it  will  then  cut  off  the 
new  Oi  axis  the  proper  length. 

This  is  an  example  of  "parallelism"  used  to 
obtain  a  proportion.  The  student  will  soon  discover 
many  further  useful  adaptations  of  the  method. 

Plate  XV.  shows  the  simple  forms  of  the  tet- 
ragonal system.  They  offer  no  difficulties  and  from 
the  figures  the  constructions  will  be  amply  plain. 


PLATE  XV. 
TETRAGONAL  SYSTEM. 

FIG.  1.  Unit  Pyramid  (1st  order)  {111} ;  c  axis  =  175. 
FIG.  2.  Unit  Pyramid  (1st  order)  {111} ;  c  axis  =  '4. 
FIG.  3.  Diametral  Pyramid  (2nd  order)  {101} ;  c  axis  = 
FIG.  4.  Ditetragonal  Pyramid  (Zirconoid)  {313}. 
FIG.  5.  Pyramid  of  the  3rd  order  {313}. 
FIG.  6.   Unit  Prism  {110}  with  base  {001}. 
FIG.  7.  Diametral  Prism  {100}  with  base  {001}. 
FIG.  8.  Ditetragonal  Prism  {310}  with  base  {001}. 


'67. 


62  HINTS  FOE  CKYSTAL  DKAWING 

The  forms  {313}  and  {313}  have  been  chosen  for 
illustration  in  Figs.  4  and  5,  merely  because  the 
intersections  are  sharper  than  in  the  forms  {212} 
{212}  they,  on  that  account,  furnish  clearer  ex- 
amples. 

Plate  XVI.  shows  the  combination  of  the  unit 
prism  with  the  unit  pyramid. 

The  example  taken,  Fig.  1,  is  zircon,  the  c  axis 
of  which,  has  the  ratio  '64  ;  that  is  to  say,  it  is  as  '64 
to  1  when  compared  with  the  horizontal  axes. 

It  has  been  thought  desirable  to  show  the  axial 
ratio  in  the  centre  of  each  figure ;  it  is  thus  con- 
venient for  reference  and  for  use  by  "  parallelism  ". 

In  Fig.  1  the  axial  ratio  is  shown  on  the  vertical 
axis,  the  point  marked  1  being  the  distance  of  the 
horizontal  semi-axis  length  from*  the  centre.  The 
point  marked  '64  shows  the  position  of  the  +  end  of 
the  c  axis. 

The  unit  prism  {110}  has  been  replaced  at  either 
end  by  a  unit  pyramid  {111}.  The  pyramid  edges 
intersect  the  prism  edges  at  a  height  less  than  that 
of  the  semi-vertical  axis. 

The  intersections  of  the  pyramid  and  prism  faces 
will  be  lines  parallel  to  lines  joining  the  ends  of  the 
«!  and  #2  axes.  The  edges  of  the  pyramid  will  be 
lines  parallel  to  lines  joining  the  ends  of  the  a  axes 
to  the  ends  of  the  c  axis. 

One  advantage  of  having  the  axial  ratios  shown 
in  the  centre  of  each  figure  is  that  edges  of  unit 
pyramids  can  at  once  be  obtained  by  "  parallelism  ". 

Fig.  2  shows  a  combination  also  of  zircon  ;  axis 
c  =  -64;  composed  of  unit  prism  {110}  and  unit 
pyramids,  but  it  is  somewhat  more  complicated  than 


(1) 


(2) 

PLATE  XVI. 
TETRAGONAL  SYSTEM. 

FIG.  1.  Zircon  c  axis  =  '64.  Prism  {110} ;  Pyramid  {111}. 
FIG.  2.  Zircon  c  axis  =  -64.  Prism  {110} ;  Pyramids  {331},  {111}. 


64  HINTS  FOK  CEYSTAL  DEAWING 

Fig.  1,  as  two  unit  pyramids  are  present.  The 
prism  gives  place  at  point  X,  a  distance  less  than  the 
semi-height  of  the  vertical  axis,  to  a  unit  pyramid 
whose  symbol  is  {331}.  This  in  its  turn  is  replaced 
by  another  pyramid,  symbol  {111}.  The  construc- 
tion will  be  sufficiently  evident  from  the  figure. 

Plate  XVII.  shows  further  combinations  in  the 
tetragonal  system.  Fig.  1  is  a  crystal  of  apophyllite  ; 
c  axis  =  1*25.  It  is  composed  of  a  prism  of  the 
second  order  {100}  and  pyramid  of  the  first  {111}. 
The  intersection  of  lines,  drawn  through  the  ends  of 
the  a  axes,  give  the  positions  for  the  vertical  edges  of 
the  prism.  The  prism  is  replaced  at  a  point  twice 
the  height  of  the  semi  c  axis  from  the  centre  by  the 
unit  pyramid. 

Through  the  point  marked  2  draw  a  line  parallel 
to  the  di  axis  to  cut  a  vertical  drawn  through  the  + 
end  of  that  axis  in  point  X ;  produce  the  line  back- 
wards and  mark  on  it  a  distance  equal  line  2  X  be- 
yond 2.  Draw  another  line  through  2  parallel  to 
the  a.2  axis  to  cut  a  vertical  through  the  +  end  of 
the  a2  axis  in  X'.  Join  points  X  and  X'.  From 
X  draw  a  line  parallel  to  one  joining  the  -h  ends  of 
the  «!  and  ,o  axes,  this  will  give  the  front  edge  of 
the  pyramid.  Bisect  the  line  joining  XXf  and 
through  the  bisection,  from  the  apex  of  the  pyra- 
mid, draw  a  line  to  cut  the  vertical  edge  of  the 
prism  in  Z.  Join  Z  to  X  and  Xf  for  intersections 
of  prism  and  pyramid  faces.  The  figure  can  then 
be  easily  finished  by  " parallelism "  and  "repeti- 
tion ". 

Fig.  2  is  a  crystal  of  rutile.  The  ratio  of  the  c 
axis  is  "64.  It  is  composed  of  prisms  of  the  first 


PLATE  XVII. 

TETRAGONAL  SYSTEM. 

FIG.  1.  Apophyllite  c  axis  =  1'25.  Prism  2nd  order  {100} ;  Unit 
Pyramid  {111}. 

FIG.  2.  Rufcile  c  axis  =  '64.  Prisms.  1st  and  2nd  order  {110}, 
{100} ;  Pyramids  1st  and  2nd  order  {111},  {101}. 

FIG.  3.  Apophyllite  c  axis  =  1-25.  Prism  2nd  order  {100};  Pyra- 
mid 1st  order  {111} ;  Basal  Pinacoid  {001}. 

FIG.  4.  Sphenoid  assumed  c  axis  =  1-5. 


66  HINTS  FOE  CEYSTAL  DEAWING 

{110}  and  second  orders  {100},  replaced  by  pyramids 
of  the  first  {111}  and  second  {101}  orders. 

The  prism  of  the  first  order  meets  the  +  end  of 
the  #2  axis  in  point  Z.  A  line  parallel  to  the  #2  axis 
through  the  -f  end  of  the  a±  axis,  cutting  a  line 
joining  the  ends  of  the  H  axes  of  the  prism  of  the 
first  order,  will  give  the  point  X  through  which  a 
vertical  edge  of  intersection  of  the  two  prisms  may 
be  drawn. 

The  prisms  are  replaced  by  a  pyramid  of  the 
second  order  at  twice  the  height  of  the  semi  c  axis 
from  the  centre. 

To  construct  this  pyramid,  draw  through  the 
point  marked  2  a  line  parallel  to  the  ^  axis,  and 
where  it  cuts  a  vertical  line  through  the  -h  end  of 
that  axis,  in  point  d,  draw  a  line  d,  u  parallel  to  the 
#2  axis  to  meet  a  line  u,  x  parallel  to  the  a^  axis 
drawn  through  a  point  t,  found  by  intersection  of 
the  line  2  t  with  a  vertical  through  the  -h  end  of  the 
a2  axis.  We  thus  have  the  base  edges  of  the  pyra- 
mid, the  portions  of  which,  that  are  intercepted 
between  the  edges  of  the  faces  of  the  prism  {100}, 
form  its  intersections  with  that  prism. 

The  apex  of  the  pyramid  may  be  found  by  "  par- 
allelism "  in  the  manner  already  described. 

Having  the  apex  we  can  draw  the  edges  of  the 
pyramid,  and  where  the  edge  u,  v  meets  a  vertical 
drawn  through  the  centre  of  the  face  of  the  prism 
of  the  first  order,  will  give  a  point  p,  to  which  we 
can  draw  the  intersections,  as  py  p',  p,  p"  of  the 
pyramid  with  that  prism. 

The  crystal  shows  also  small  faces  of  a  unit 
pyramid  ;  this  meets  the  line  d  2  in  a  point  e. 


TETEAGONAL  SYSTEM  67 

The  intersecting  edge  of  the  two  pyramids  may 
be  found  by  drawing  the  line  e,  g,  which  is  parallel 
to  that  joining  the  4-  ends  of  the  a^  and  a2  axes.  This 
line  will  cut  the  base  edge  of  the  pyramid  of  the 
second  order  in  <?,  and  from  the  point  where  it  cuts, 
a  line  may  be  drawn  parallel  to  the  front  edge  of 
the  pyramid  {111}  marked  g,  h.  This  line  will  cut 
the  edge  of  the  second  order  pyramid  and  give  the 
intersection  of  the  faces  of  the  two  pyramids. 

The  intersection  of  the  unit  pyramid  with  the 
unit  prism  i,  k,  will  be  parallel  to  the  line  joining 
the  4-  ends  of  the  0,1  and  a2  axes. 

These  points  found,  the  crystal  can  easily  be 
completed  by  "  parallelism  "  and  "  repetition  ". 

Fig.  3  is  a  crystal  of  apophyllite ;  c  axis  =  1  "25, 
consisting  of  a  prism  of  the  second  order  {100},  a 
pyramid  of  the  first  order  {111}  and  the  basal  pina- 
coid  {001}. 

The  basal  pinacoid  cuts  the  prism  at  a  point  H. 

A  line  drawn  through  point  H  parallel  to  the 
a2  axis  to  cut  a  vertical  through  the  4-  end  of  that 
axis  will  give  a  point  through  which  to  draw  the 
intersection  of  base  and  prism. 

A  face  of  the  pyramid  cuts  the  intersection  of 
prism  and  base  in  point  x. 

The  intersection  of  pyramid  and  base  will  be  a 
line  parallel  to  one  joining  the  4-  ends  of  the  //axes. 

The  intersection  of  pyramid  with  prism  will  be 
parallel  to  the  edges  of  the  pyramid.  The  inter- 
sections with  the  front  face  of  the  prism,  being 
parallel  to  the  side  edges  of  the  pyramid,  those  with 
the  side  faces  of  prism,  to  the  front  and  back  edges 
of  the  pyramid.  The  reason  is  evident.  The  front 


68  HINTS  FOE  CKYSTAL  DBA  WING 

face  of  the  prism  being  parallel  to  the  vertical 
axial  plane  through  the  a2  and  c  axes  will  show 
parallel  intersections. 

Fig.  4  shows  the  form  called  a  sphenoid  {ill}- 
The  axial  ratio  for  the  c  axis  has  been  taken  as  1  '5. 
The  edges,  as  in  the  similar  form  of  the  isometric 
system,  the  tetrahedron,  are  parallel  to  lines,  edges 
of  the  pyramid  {111}- 

Plate  XVIII.  shows  a  somewhat  complicated 
crystal  of  zircon,  the  ratio  of  the  c  axis  being  -64. 

The  forms  present  are  a  prism  of  the  first  order, 
the  edges  of  which  are  truncated  by  one  of  the 
second  order,  whose  faces  intersect  the  a  axes  of 
the  first  prism  at  points  p  and  p'. 

The  ends  of  the  crystal  are  formed  by  pyramids, 
the  pyramid  {111},  prevailing.  Small  faces  of  the 
form  {311}  cut  off  the  angles  at  the  ends  of  the 
horizontal  axes  of  the  first  pyramid. 

The  unit  pyramid  {111}  replaces  the  prisms  at  a 
point  g. 

The  construction  of  the  prisms  and  unit  pyramid 
{111},  should  now  offer  no  difficulty,  but  the  inter- 
sections of  these  with  the  ditetragonal  pyramid  {311} 
need  explanation. 

The  ditetragonal  pyramid  cuts  the  prism  of  the 
second  order  in  the  point  S,  the  point  where  the 
prism  face  meets  the  di  axis  of  the  unit  pyramid. 

An  O  plan  of  the  axes  of  the  unit  pyramid  is 
constructed  above  the  figure,  a  vertical  taken  up- 
wards from  point  S  of  the  C  projection  will  cut  the 
axis  DC  of  the  plan  in  the  point  S'  where  the  H 
trace  of  the  ditetragonal  pyramid  meets  it. 

Lines  SX X  on  the  O  plan  are  the  H  traces  of 


PLATE  XVIII. 

TETRAGONAL  SYSTEM. 

Zircon  c  axis  =  -64.     Prisms  of  1st  {110}  and  2nd  {100}  orders  ;  Unit 
Pyramid  {111}  and  Ditetragonal  Pyramid  {311}. 


70  HINTS  FOR  CRYSTAL  DRAWING 

two  faces  of  the  ditetragonal  pyramid.  They  cross 
the  H  trace  of  the  unit  pyramid  in  point  0.  Another 
point  where  the  edges  of  the  two  pyramids  intersect 
is  found  by  the  "  profile  "  method  thus  : — 

The  right-angled  triangle  CC'D  is  the  profile 
cut  through  the  front  edge  of  the  unit  pyramid  and 
turned  down  on  line  DC  as  hinge  line. 

The  right-angled  triangle  CSB  is  the  profile  of 
the  ditetragonal  pyramid  {311}  cut  through  its  front 
edge  and  turned  down  on  line  CS  as  a  hinge 
line. 

E  is  the  point  where  the  two  front  edges  of  the 
two  pyramids  cut  each  other  ;  a  line  drawn  from  E 
at  right  angles  to  CD  gives  a  point  immediately 
below  E,  point  E'.  E  is  in  fact  the  plan  of  point 
E  on  the  0  plan. 

Dropped  by  a  vertical  projector  to  the  C  pro- 
jection it  cuts  the  front  edge  of  the  unit  pyramid  at 
its  point  of  intersection  with  the  front  edge  of  the 
ditetragonal  pyramid  in  n. 

Having  now  two  points  n,  n',  common  to  the 
faces  of  both  pyramids,  we  can  join  them  for  the 
edge  of  intersection. 

The  intersections  of  the  prism  {100}  and  pyramid 
{311}  are  easily  found  ;  they  pass  through  ri  and  the 
corresponding  points ;  they  are  parallel  to  lines 
joining  the  ends  of  the  a2  and  c  axes  for  the  back 
and  front  faces — to  lines  joining  the  ends  of  the  #x 
and  c  for  the  side  faces.  This  will  seem  a  little 
strange ;  careful  consideration  of  the  indices  of  the 
symbols  involved  will  make  it  clear.  The  inter- 
sections are  parallel  to  those  of  the  particular 
pyramid  face  to  which  they  belong,  with  that  axial 


TETKAGONAL  SYSTEM  71 

plane,  with  which  the  prism  face  they  intersect  is 
parallel. 

Exercises — Problems  for  solution. 
Tetragonal  System — 
Kutile  c  axis  =  0*64.     {110} ;  {100};  {111}- 


CHAPTEK  X. 

HEXAGONAL  SYSTEM— CONSTRUCTION  FOR  AXES— SIMPLE 
FORMS— COMBINATIONS. 

PLATES  XIX.  TO  XXIII. 

IN  the  hexagonal  system  there  are  four  axes.  Three 
of  these  are  horizontal,  equal,  and  make  with  each 
other  angles  of  60°.  They  are  lettered  respectively 
al9  «2,  ay 

The  fourth  axis  is  vertical,  therefore  at  right 
angles  to  the  other  three,  its  length  is  variable.  It 
is  lettered  c. 

Plate  XIX.  shows  the  method  of  finding  the 
hexagonal  axes  in  C  projection. 

As  for  the  isometric  and  tetragonal  systems,  we 
begin  by  making  an  0  plan  of  the  axes  above. 

Only  the  half  plan  is  required.  We  construct 
it  thus  ;— 

Draw  the  horizontal  line  XY.  Select  a  point 
c  for  centre  and  draw  line  c,  +  A2,  making  the 
angle  18°  26'  with  XY ';  cut  it  off  the  length  desired 
for  the  horizontal  axes.  It  is  the  plan  of  the  +  half 
of  the  az  axis. 

To  get  the  +  half  of  the  a^  and  -  half  of  the  as 
axes,  we  describe  a  circle  with  c,  +  A2  as  radius, 
from  c  as  centre,  then  with  the  same  radius  and 
the  4-  end  of  the  A2  axis  as  centre,  we  cut  the 
circle  in  a  point  which  will  be  the  —  end  of  the 

72 


0'  ft  it.  * 


c      0"  * 


Profile. 

(2) 


PLATE  XIX. 
HEXAGONAL  SYSTEM. 
PIG.  1.  Construction  of  axes. 
FIG.  2.  Profile  showing  apparent  depression. 


74  HINTS  FOB  CKYSTAL  DBA  WING 

A 3  axis,  join  the  point  to  c  for  the  -  half  of  the 
A?  axis. 

With  the  same  radius,  using  the  -  end  of  the  As 
axis  as  centre,  cut  the  circle  again ;  this  will  give 
a  point  for  the  +  end  of  the  a^  axis,  join  it  to  c 
for  the  half  of  the  axis. 

We  obtain  the  C  projection  from  the  0  plan  in 
exactly  the  same  way  as  for  the  two  former  systems 
by  dropping  verticals  from  c  and  from  the  ends  of 
the  axes  of  the  0  plan.  The  apparent  depressions 
for  the  several  axes  are  got  by  taking  \  distances 
between  On,  O'n'  and  0"V  and  marking  them 
down  from  X'  Yf  on  the  respective  verticals. 

The  vertical  c  axis  is  found  just  as  in  the  tetra- 
gonal system  by  finding  its  ratio,  the  horizontal  axes 
being  taken  as  unity.  The  ratio  for  beryl  '8  has 
been  used  in  the  illustration. 

The  student  should  carefully  note  the  axial 
lettering  and  bear  in  mind  that  the  near  end  of  as 
axis  is  the  minus  end. 

Plate  XX.  gives  the  simple  forms  of  the  hexa- 
gonal system.  It  needs  no  comment.  The  method 
of  finding  the  horizontal  pyramidal  edges  of  Fig.  2, 
the  pyramid  of  the  second  order,  from  the  0  plan 
may  be  noted.  Vertical  projectors  are  dropped  to 
cut  lines  drawn  through  the  bisection  of  the  angles 
made  by  the  axes. 

Plate  XXI.  Fig.  1  shows  the  construction  for 
the  dihexagonal  pyramid  {2133}  and  the  pyramid  of 
the  third  order  Fig.  2.  Both  are  worked  from  the 
0  plan  because  it  gives  sharper  intersections  than 
plane  intersection  of  the  C  projection.  Propor- 
tional points  have  been  used. 


PLATE  XX. 
HEXAGONAL  SYSTEM. 

FIG.  1.  Pyramid  of  the  1st  order  {1011}. 
FIG.  2.  Pyramid  of  the  2nd  order  [1122}. 
FIG.  3.  Prism  of  the  1st  order  {1010}. 
FIG.  4.  Prism  of  the  2nd  order  {Il20}. 


76  HINTS  FOR  CRYSTAL  DRAWING 

Plate  XXII.  shows  a  somewhat  complicated 
crystal  of  iodyrite,  c  axis  ratio  =  '8,  belonging  to  a 
lower  class  of  symmetry  in  which  half  the  faces  are 
suppressed.  It  is  a  combination  of  a  prism  of  the 
second  order  {1150}  and  two  pyramids  of  the  first 
order  {4041}  and  {4045}- 

The  figure  illustrates  how  few  lines  are  really 
essential  for  construction. 

The  prism  of  the  second  order  is  first  found  in 
the  usual  way  by  verticals  through  the  ends  of  the 
horizontal  axes.  The  prism  is  replaced  at  a  dis- 
tance TS  from  the  centre  by  a  unit  pyramid  having 
the  symbol  {4041}- 

The  apex  of  this  pyramid  might  be  found  by 
marking  off  the  c  axial  ratio  four  times  on  the 
vertical  from  point  0,  but  this  entails  making  the 
figure  unduly  high,  so  instead  we  may  find  it  by 
proportion  thus  : — 

Draw  line  OO  parallel  to  the  semi  a2  axis.  Join 
the  J  point  of  this  line  to  H  the  height  of  the  C  axis 
marked  from  0.  Then  a  line  from  0'  parallel  to  the 
one  from  the  \  point  to  H,  will  give  the  inclination 
for  the  right-hand  edge  of  the  pyramid  without  it 
being  necessary  to  have  the  apex  on  the  paper. 

The  pyramid  is  cut  off  at  a  level  of  three  times 
the  semi- vertical  axis  by  the  basal  plane  {0001}- 

A  line  drawn  through  the  centre  parallel  to  the 
az  axis  at  this  level  will  cut  the  edge  of  the  pyramid 
already  drawn,  in  L.  The  remaining  pyramidal 
edges  may  be  found  by  lines  in  the  basal  plane 
parallel  to  the  three  horizontal  axes,  intersected  by 
lines  parallel  to  those  joining  the  ends  of  the  hori- 
zontal axes.  The  intersections  give  us  the  points 


78  HINTS  FOR  CRYSTAL  DRAWING 

I'  I"  l'»  lfm  I"'"  on  the  basal  plane  to  which  the 
several  pyramidal  edges  can  be  drawn. 

Even  when  the  basal  plane  is  not  present  in  a 
crystal,  it  is  clear  that  a  horizontal  plane  may  be 
assumed  and  thus  the  pyramidal  edges  found  with- 
out using  the  apex. 

We  have  next  to  obtain  the  intersections  of 
prismatic  and  pyramidal  faces ;  proceed  as  fol- 
lows :— 

Draw  on  the  O  plan,  used  for  constructing  the 
axes,  the  H  traces  of  the  prism,  they  will  of  course 
be  lines  at  right  angles  to  the  axes.  Drop  a  ver- 
tical from  point  X,  the  intersection  of  two  of  the 
H  traces ;  cut  the  vertical  in  a  point  P  by  a  line 
passed  through  the  centre  of  line  SS'  and  the  centre 
of  the  basal  edge  parallel  to  it,  VI".  Point  P  is 
one  of  the  points  of  intersection  which  may  be 
joined  to  S  and  Sf,  two  other  such  points ;  thus  we 
get  two  edges  of  intersection. 

The  other  intersection  edges  of  prism  and  pyra- 
mid may  be  found  by  drawing  from  P,  line  PP' 
parallel  to  NN'.  Make  the  distance  P'Y  =  PY. 
Point  P"  and  corresponding  points,  may  be  found 
similarly. 

We  now  have  only  to  construct  the  unit  pyramid 
at  the  lower  end  to  complete  the  crystal. 

The  symbol  for  this  form  is  {4045}-  It  replaces 
the  prism  at  the  same  distance  from  the  centre  be- 
low, as  the  other  pyramid  above.  The  height  then 
of  its  semi  c  axis,  will  be  not  '8  of  line  CD  as  for 
the  prism,  but  *8  of  I  of  CD,  since  the  axial  ratio  of 
the  pyramid  is  {4045} ;  that  is,  1-25  :  0  :  1*25  : 1. 

From  point  R,  found  by  making  TR  equal  ST, 


6  FOR  LOW£R  PYRAMID 

8D 


PLATE  XXII. 

HEXAGONAL  SYSTEM. 


lodyrite  c  axis  =  *8. 

Prism  of  the  2nd  order  {1120} ;  Pyramid  1st  order  {4041}. 

Pyramid  1st  order  {40i1>} ;  Base  {0001}. 


80  HINTS  FOE  CBYSTAL  DBA  WING 

draw  a  line  parallel  to  the  #3  axis  and  from  its  in- 
tersection with  the  vertical  through  the  centre  of  the 
prism,  point  Z7,  mark  downwards  a  length  equal  the 
semi  c  axis  ;  for  the  lower  pyramid  -8,  shown  on  the 
0  plan,  this  will  be  the  apex  of  the  pyramid.  The 
pyramidal  edges  join  the  points  of  the  several  faces 
of  the  prism  correspondent  to  point  R,  to  the  apex 
of  the  pyramid. 

To  get  the  intersections  of  pyramidal  and  pris- 
matic faces,  join  R  to  the  next  correspondent  point, 
bisect  the  joining  line  and  pass  a  line  through  the 
bisection  and  the  apex  to  cut  the  line  PQ  in  p'. 
R  joined  to  p'  is  one  edge  of  intersection.  The 
others  may  be  found  by  "  repetition  ". 

The  reasons  for  the  several  steps  taken  in  con- 
structing the  figure  should  be  apparent  without 
further  explanation. 

Plate  XXIII.  shows  a  crystal  of  connelite.  All 
the  construction  lines  are  not  shown  in  the  figure. 

The  0  plan  gives  the  H  traces  of  the  prism  of 
the  first  order  and  of  the  portions  of  the  faces  of 
the  prism  of  the  second  order  {1120}  which  truncate 
those  of  the  prism  of  the  first  order  {IrtlO}-  The 
points  of  intersection  are  dropped  to  the  C  projec- 
tion below. 

The  prisms  are  replaced  by  a  berylloid  or  dihex- 
agonal  pyramid  11  :  2  :  13  :  3  at  a  point  X.  Through 
this  point  the  centre  lines  parallel  to  the  H  axes  and 
lines  bisecting  the  angles  between  the  axes  are 
drawn. 

The  proportional  points  11:2  are  found  on  the  a1 
and  #2  axes  of  the  0  plan,  and  the  horizontal  traces 
g  i  h  drawn  by  "  parallelism,"  i  h  being  parallel  to 


PLATE  XXIII. 
HEXAGONAL  SYSTEM. 

Connelite  c  axis  =  1'16. 
Prisms  1st  order  {Ill0},.2nd  {1120}. 

Pyramid  1st  order   {llll}  ;    Dihexagonal   Pyramid   (Berylloid) 
Axial  Ratio  11 :  2  : 13  : 3. 


82  HINTS  FOE  CEYSTAL  DEA^VING 

the  line  from  points  found  in  proper  proportion  on 
the  axes. 

g  i  h  are  dropped  to  the  C  projection  in  gf  ir  h', 
and  the  other  correspondent  points  are  found  by 
"  repetition  ". 

The  apex  of  the  berylloid  is  marked  at  A,  a  point 
4  times  and  \  the  semi- vertical  axis  from  point  Jf, 
in  accordance  with  the  ratio  11  :  2  :  13  :  3,  and  to- 
wards this  point  A,  lines  from  the  points  g'hfi', 
etc.,  are  directed  to  form  the  berylloid  edges. 

The  intersections  of  berylloid  and  prism  faces  is 
the  next  problem. 

Vertical  lines  through  the  centres  of  the  prism 
faces,  drawn  upwards  to  cut  the  pyramidal  edges  at 
dd'  give  points  to  which  the  prism  and  pyramid  in- 
tersections may  be  drawn  from  the  ends  of  the  prism 
edges  at  e,  etc. 

The  berylloid  in  its  turn  is  replaced  by  a  unit 
pyramid  {llll}  at  point  U. 

Through  U  draw  lines  parallel  to  the  H  axes 
to  cut  the  berylloid  edges.  Find  the  apex  of  the 
pyramid  and  join  the  points  of  cutting  to  it. 

To  find  the  intersections  of  berylloid  and  pyra- 
midal faces,  draw  the  H  edges  of  the  pyramid,  bisect 
them  and  draw  lines  from  the  apex  through  the 
bisections,  producing  them  downwards  to  cut  the 
berylloid  edges. 

The  points  so  found  must  be  joined  to  the  meet- 
ing points  of  pyramidal  and  berylloid  edges ;  the 
joining  lines  will  be  the  desired  intersections. 

The  figure  can  now  be  achieved  by  " parallelism". 


CHAPTEK  XL 

HEXAGONAL  SYSTEM  (cotrr.)— RHOMBOHEDRON— SCALENO- 
HEDRON— TRAPEZOHEDRON. 

PLATES  XXIV.  TO  XXVII. 

PLATE  XXIV.  Fig.  1  is  the  +  rhombohedron 
{1011}  of  hematite  c  axis  =  1*36.  Its  faces  cor- 
respond to  alternate  faces  of  the  unit  pyramid. 

Having  drawn  the  axes  by  the  construction 
given  on  Plate  XIX.,  join  the  4-  end  of  the  a\  axis 
to  the  -  end  of  the  aa,  producing  the  line  on  either 
side.  Also  join  the  —  end  of  the  a2  to  the  +  end 
of  the  #3  axis,  producing  the  line  to  the  left  to  cut 
the  first  drawn  line. 

These  two  lines  are  the  H  traces  of  planes  {1011} 
and  {Olll}-  Where  they  cut  is  a  point  common  to 
both  planes ;  the  apex  of  the  vertical  axis  is  another 
point  common  to  both.  Join  these  two  common 
points,  the  joining  lines  are  the  intersections  of  the 
planes  forming  two  faces  of  the  rhombohedron. 
Join  the  -f  end  of  the  a±  axis  to  the  -  end  of  the  ay 
Bisect  the  line  joining  them ;  through  the  point  of 
bisection  and  the  -  end  of  the  c  axis  draw  a  line  to 
cut  the  intersection  of  the  planes  in  point  i. 

Now  find  in  a  similar  way  the  intersection  of 
planes  UOll}  and  {1101}  by  the  intersection  of  the 
ff  traces,  giving  a  point  common  to  both  planes, 
another  point  common  to  both  being  the  apex  of  the 

83  6* 


84  HINTS  FOE  CEYSTAL  DEAWING 

c  axis.     Cut  off  the  intersection  of  the  planes  in 
point  s  by  a  line  from  the  —  end  of  the  c  axis  drawn 
through  the  bisection  of  the  line  joining  the  —  az  and 
.  -f  a2  axes. 

.    From  s  draw  a  line  st  t  parallel  to  i,  v  and  make 
it  an  equal  length  on  each  side  of  the  -  az  axis. 

Line  ts  is  another  edge  of  the  rhombohedron 
which  may  be  easily  completed  by  "parallelism" 
and  "  repetition  ". 

Fig.  2  shows  a  -  rhombohedron  {0111}  of  cal- 
cite  c  axis  =  *85.  It  is  constructed  similarly  to  the 
+  rhombohedron,  the  alternate  planes  omitted  in 
the  +  form  being  now  developed. 

Fig.  3  shows  a  combination  of  the  +  rhombohe- 
dron UOll}  with  the  base  {0001}- 

The  basal  planes  cut  the  vertical  axis  in  points 
XX1.  The  intersections  with  the  rhombohedral 
planes  are  found  by  joining  the  axis  through  which 
the  edge  of  the  rhombohedron  passes  to  the  end  of 
the  c  axis,  then  passing  through  X  or  X'  a  line 
parallel  to  the  H  axis  to  cut  the  line  going  to  the  c 
axis.  The  place  where  they  cut,  is  a  point  through 
which  may  be  drawn  the  intersection  of  the  rhom- 
bohedral face  with  the  basal  plane.  The  several 
intersections  will  be  parallel  to  the  several  axes. 

Fig.  4  shows  a  -  rhombohedron  {0221}  of 
hematite  c  axis  =  1*36,  whose  edges  are  truncated 
by  the  +  rhombohedron  {1011}- 

To  understand  the  relation  of  the  truncated 
rhombohedron  to  the  other,  is  of  great  assistance  in 
the  construction  of  this  combination.  It  is  this,  the 
+  rhombohedron  to  truncate  the  minus  form,  has 
twice  the  latter's  length  of  vertical  axis. 


(2) 

PLATE  XXIV. 

HEXAGONAL  SYSTEM. 

FIG.  1.  Hematite  c  axis  =  1'36 ;   +  Rhombohedron  {lOll}. 

FIG.  2.  Calcite  c  axis  =  '86  ;    -  Rhombohedron  {0111}. 

FIG.  3.  Calcite  +  Rhombohedron  cut  by  Basal  Plane. 

FIG.  4.    f  Rhombohedron  {1011}  truncating  edges  of  -  Rhombohedron  {022l}. 


86  HINTS  FOE  CEYSTAL  DEAWING 

Having  found  the  point  on  the  vertical  axis,  in 
this  case  75  from  the  centre,  through  which  the 
truncating  face  passes,  we  merely  have  to  draw  a 
line  p,  p'  parallel  to  the  edge  of  the  -  rhombohedron 
to  cut  a  line  p'x  through  the  centre  of  the  lower 
face,  then  through  the  point  of  cutting  we  can  draw 
an  intersection  parallel  to  the  a2  axis ;  where  this 
meets  the  edges  of  the  -  rhombohedron,  lines 
parallel  to  p,  pr  will  give  us  other  intersections. 
The  figure  may  be  completed  by  "  repetition  ".  The 
apex  of  the  combination  being  where  line  ppf  meets 
the  vertical  axis. 

Plate  XXV.  illustrates  the  combination  of  the 
rhombohedron  with  the  prism. 

Fig.  1  gives  the  combination  of  the  unit-prism 
with  -  rhombohedron  {0112},  in  a  crystal  of  calcite, 
G  axis  =  *85. 

Through  point  P,  the  point  on  the  V  axis  where 
the  rhombohedron  replaces  the  prism,  draw  lines 
parallel  to  the  H  axes  to  cut  the  prism  edges  in 
points  d,  e,  /,  g,  h,  i.  Join  d-e,  f-g,  h-i,  for  the  inter- 
section of  prismatic  and  rhombohedral  edges.  Pro- 
duce the  intersections  on  either  side  until  they  cut. 
Join  the  point  where  they  cut  to  one  half  the  height 
of  the  semi-vertical  axis  in  accordance  with  the 
symbol  {0112}-  Cut  off  the  lines  so  drawn  by 
verticals  through  the  centres  of  the  prism  faces 
erected  through  the  bisection  points  of  lines  e-ft  g-h, 
etc. 

We  then  have  all  the  needed  intersection  edges 
and  can  complete  the  figure  as  shown,  treating  the 
lower  end  of  the  crystal  in  a  similar  manner. 

Fig.  2  is  a  crystal  of  dioptase,  vertical  axis  =  '53. 


PLATE  XXV. 

HEXAGONAL  SYSTEM. 

FIG.  1.  Calcite  c  axis  =  -85.  Unit  Prism  {UlO}  and  -  Rhombo- 
hedron  {0112}. 

FIG.  2.fcDiopfcase  c  axis  =  '53.  Prism  2nd  order  {1120}  and  - 
Rhombohedron_022l}. 


88  HINTS  FOE  CEYSTAL  DEAWING 

It  shows  a  combination  of  prism  of  the  second  order 
with  the  -  rhombohedron  {0221}- 

Draw  the  prism  faces  first  with  the  aid  of  the 
portion  of  the  0  plan  above,  dropping  vertical  pro- 
jectors from  XX',  the  H  traces  of  the  prism  faces, 
to  cut  lines  drawn  in  the  C  projection  at  right  angles 
to  the  H  axes.  Only  one  of  these  lines  is  shown  at 
X". 

As  in  the  last  figure,  through  the  point  P  where 
the  rhombohedron  replaces  the  prism,  draw  lines 
parallel  to  the  several  H  axes  of  the  crystal ;  these 
will  give  the  /faxes  of  the  rhombohedron.  Join  the 
ends  of  the  -h  di  and  -  az  axes ;  bisect  the  joining 
line. 

Next  find  point  L  twice  the  height  of  the  semi- 
vertical  axes  from  P  in  accordance  with  the  symbol 
{0221}-  Draw  lines  from  this  point  L  through  the 
bisections  of  the  lines  joining  the  H  axes  of  the 
rhombohedron ;  produce  them  to  cut  the  vertical 
edges  of  the  prism  at  k,  /,  n.  Oblique  lines  drawn 
from  k,  /,  ft,  through  the  ends  of  the  H  axes  of  the 
rhombohedron  and  produced  to  cut  the  vertical  edges 
of  the  prism  in  v,  q,  f,  will  give  the  other  points 
required  for  the  intersection  of  the  rhombohedral 
and  prismatic  faces. 

It  is  well  to  test  the  accuracy  of  this  latter  step 
by  marking  the  distance  from  P  to  Ly  downwards  in 
S,  then  from  S  through  the  central  points  of  the 
lines  joining  the  ends  of  the  H  axes  draw  lines  to 
cut  the  edges  from  L.  They  should  cut  in  the  points 
q,  r,  o.  One  such  line  is  shown  for  example,  line  q,  s. 

The  lower  portion  of  the  crystal  can  be  got  by 
"  parallelism  "  and  "  repetition  ". 


PLATE  XXVI. 

HEXAGONAL  SYSTEM. 
Calcite  c  axis  =  '85  ;   +  Scalenohedron  {2131}. 


90  HINTS  FOE  CKYSTAL  DEAWING 

Plate  XXVI.  shows  the  scalenohedron,  calcite 
G  axis  =  -85  {2131},  +  form. 

The  construction  is  based  on  that  of  the  unit  + 
rhombohedron  {1011}- 

First  find  the  rhombohedral  edges  by  the  method 
given  (Plate  XXIV.  Fig.  1).  Then  mark  off  the 
vertical  axis,  in  this  case  three  times  the  axial  ratio 
of  calcite.  Join  the  ends  of  each  rhombohedral  edge 
to  the  apex  ;  the  form  is  then  complete. 

Plate  XXVII.  gives  a  -h  trapezohedron  {2131} 
of  quartz,  the  c  axis  =  1*1. 

Start  as  for  the  scalenohedron  with  the  rhom- 
bohedral edges  but  retain  and  produce  only  alternate 
ones.  Find  points  e,  f,  g,  the  intersections  of  the  H 
traces  of  planes  {2131}  with  {3211}  and  of  {2131} 
with  {1321}-  Join  the  points  of  intersection  to  the 
apex.  The  lines  going  to  the  apex  will  cut  off  the 
rhombohedral  edges  the  right  length  on  one  side. 
Mark  off  the  length  on  the  other  side  of  the  point 
where  the  axis  meets  the  edge,  thus  completing  the 
form  by  lines  joining  the  ends  of  the  edges. 

Exercises — Problems  for  solution. 

Hexagonal  System — 

Hematite    caxis  =  l'36.  {0001};  {1010}- 

Quartz        caxis  =  l'09.  {1011};  {0111};  {1010}- 

Calcite        caxis=   -85.  {lOll}- 

Apatite       caxis=    73.  {1010};  {1011};  {0001}- 

Nepheline  caxis=    '84.  {1010};  {OOOl}- 


PLATE  XXVII. 

HEXAGONAL  SYSTEM. 
Quartz  c  axis  =  11 ;  Eight-handed  Trapezohedron  {2131}. 


CHAPTEK  XII. 

ORTHORHOMBIC  SYSTEM— AXES— SIMPLE  FORMS— COMBINA- 
TIONS. 

PLATES  XXVIII.  TO  XXXI. 

THE  axes  of  the  orthorhombic  system  are  all  at  right 
angles  but  of  different  lengths,  they  therefore  are  all 
lettered  differently.  The  axis  correspondent  to  the 
«i  is  lettered  a,  that  correspondent  to  the  az  is  b,  the 
vertical  axis  is  c. 

The  a  axis  is  the  shorter  of  the  horizontal  axes 
and  is  sometimes  called  the  brachy  axis,  from  the 
Greek  /Spa^us,  short. 

The  b  axis  is  sometimes  called  the  macro  axis, 
from  the  Greek  /x,aK/>ds,  long. 

The  construction  for  the  axes  is  precisely  that 
for  the  isometric  and  tetragonal  systems,  the  only 
difference  being  that  the  H  axes  as  well  as  the  V 
axis  are  made  of  a  length  correspondent  to  the 
ratio  characteristic  of  the  crystal  to  be  drawn,  the  b 
axis  being  taken  as  unity. 

Plates  XXVIII.  and  XXIX.  give  the  simple 
forms. 

On  Plate  XXVIII.  Fig.  1  is  shown  the  unit 
prism  {110}  with  the  basal  pinacoid  {001  }•  Fig.  .2 
illustrates  the  relations  of  the  macropinacoid  {100}, 
macroprism  {210},  unit  prism  {110}>  brachyprism 
{120}  and  basal  pinacoid  {001}- 

92 


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94  HINTS  FOE  CKYSTAL  DBA  WING 

The  figure  has  preferably  been  constructed  from 
the  0  plan,  a  portion  of  which  is  shown  above,  rather 
than  by  plane  intersection  directly  on  the  C  projec- 
tion, because  on  the  0  plan  the  lines  cut  each  other 
more  acutely. 

Plate  XXIX.  Fig.  1  shows  the  combination  of 
the  three  pinacoids  :  the  macropinacoid  parallel  to 
the  macro  axis  {100}»  the  brachypinacoid  parallel  to 
the  brachy  axis  {010}  and  the  basal  pinacoid  parallel 
to  both  these  axes  {001}- 

Fig.  2  shows  the  unit  pyramid  {111};  the  con- 
struction needs  no  comment. 

Fig.  3  shows  the  macrodome  {101}  in  combina- 
tion with  the  brachypinacoid  {010}- 

Fig.  4  shows  the  brachydome  {011}  combined 
with  the  macropinacoid  {100}- 

Plate  XXX.  shows  several  combinations  in  the 
orthorhombic  system.  Fig.  1,  a  crystal  of  barite 
axial  ratio  '8  : 1 :  1  '3,  is  composed  of  a  unit  prism 
{110}  and  macrodome  {102}. 

The  face  of  the  macrodome  meets  the  a  axis  of 
the  prism  at  point  X. 

Join  the  +  ends  of  the  a  and  b  axes,  erect  the 
vertical  edges  of  the  prism  through  the  ends  of  the  b 
axis.  Draw  the  intersection  edge  of  the  macrodome 
faces  through  point  X  parallel  to  the  b  axis.  Where 
the  edge  of  intersection  meets  the  line  joining  the 
ends  of  the  a  and  b  axes  in  N  will  be  one  point 
where  the  macrodomal  and  prismatic  faces  meet. 

Draw  a  line  from  the  +  end  of  the  a  axis  to  \  the 
height  of  the  semi  c  axis,  then  a  line  parallel  to  it 
from  point  X  to  cut  that  axis  in  point  P.  Line  XP 
is  the  line  of  inclination  of  the  macrodome  face. 


001 


100 


010 


(1) 


PLATE  XXIX. 

ORTHORHOMBIC  SYSTEM. 

FIG.  1.  Macropinacoid  {100} ;  Brachypinacoid  {010} ;  Basal  Pina- 
coid  {001}. 

FIG.  2.  Unit  Pyramid  {111}. 

FIG.  3.  Macrodome  {101} ;  Brachypinacoid  {010}. 

FIG.  4.  Brachydome  {011} ;  Macropinacoid  {100}. 


96  HINTS  FOE  CKYSTAL  DEAWING 

Through  P  the  edge  of  macrodome  may  now  be 
drawn  parallel  to  the  b  axis  ;  it  will  cut  the  vertical 
edge  of  the  prism  at  V.  V  then  will  be  a  point 
common  to  both  macrodomal  and  prismatic  faces,  so 
that  by  joining  V  and  N  we  obtain  the  intersection 
of  those  faces. 

Complete  the  crystal  by  "parallelism"  and  "  re- 
petition ". 

Fig.  2  is  another  form  of  barite.  In  addition  to 
the  unit  prism  and  small  faces  of  macrodome  shown 
at  d,  etc.,  the  brachydome  O  {011}  and  basal  pina- 
coid  {001}  are  present. 

First  draw  the  unit  prism,  then  through  point  B 
the  point  on  the  vertical  axis  through  which  the 
basal  plane  passes,  draw  lines  parallel  to  the  a  and 
b  axes  to  cut  off  the  vertical  prism  edges. 

For  the  macrodome  face  join  the  +  end  of  the  a 
axis  to  %  the  semi  height  of  the  c  axis,  draw  a  line 
parallel  to  this  line  through  point  e,  where  the  mac- 
rodome face  cuts  the  prism  edge.  Through  the  point 
where  the  line  from  e  meets  the  parallel  to  the  a 
axis  from  B  draw  the  intersection  of  basal  pinacoid 
and  macrodome,  parallel  to  the  b  axis,  to  cut  the  H 
edges  of  intersection  of  basal  plane  with  prism  faces, 
in  points  i,  if,  then  join  these  points  of  cutting  to  e  for 
the  intersections  of  macrodomal  and  prism  faces. 

To  obtain  the  brachydome  faces  join  the  ends  of 
the  b  axis  to  the  ends  of  the  c  axis,  this  gives  the 
lines  of  inclination  of  the  brachydomal  faces  ;  where 
they  cut  the  parallel  to  the  b  axis  through  B  are 
points  t,  t',  through  which  the  intersections  of  brachy- 
dome and  base  may  be  drawn,  parallel  to  the  a  axis 
to  cut  the  //intersections  of  base  and  prism  in  points 


dom6 


PLATE  XXX. 

ORTHORHOMBIC  SYSTEM. 
Barifce<     Axial  Rat*0  *8  :  1  :  1-3  ;  Unit  Prism  {110}  ;  Macro- 


F??'r?A  ?rookite-    Axial  Ratio  '8  :  1  :  «9  ;  Unit  Prism  {110}  ;  Brachy- 
pyramid  {122}. 

FIG.  4.  Arsenopyrite.    Axial  Ratio  "67  :  1  :  118  ;  Unit  Prism  {110}  ; 
Brachydome  {Oil}. 

FIG.  5.  Arsenopyrite.     Unit  Prism  {110}  ;  Brachydome  {014}. 


98  HINTS  FOE  CEYSTAL  DEAWING 

k,  kf,  I,  I'.  Join  the  points  of  cutting  to  the  ends  of 
the  b  axis  for  the  intersections  of  brachydomal  and 
prismatic  faces. 

Fig.  3  is  a  crystal  of  brookite,  axial  ratio  -8:1: 
•9.  It  is  formed  by  unit  prism  {110}  and  brachy- 
pyramid  {122}-  The  construction  is  so  simple  it  will 
need  no  comment. 

Figs.  4  and  5  are  crystals  of  arsenopyrite,  axial 
ratio  -67  : 1 :  1  18. 

They  are  both  combinations  of  brachydomes  with 
unit  prism;  the  brachydome  in  Fig.  4  having  the 
symbol  {Oil},  that  in  Fig.  5  the  symbol  {014}-  The 
brachydomes  both  cut  the  vertical  edge  through  the 
end  of  the  b  axis  at  a  point  a  little  above  the  centre 
of  the  crystal.  The  construction  will  be  evident  at 
sight. 

Plate  XXXI.  shows  further  combinations  in  the 
orthorhombic  system.  Fig.  1  represents  a  crystal 
of  childrenite,  axial  ratio  77 : 1 :  '53,  composed 
of  macropinacoid  {100}>  brachypinacoid  {010}  and 
pyramid  {121}- 

For  the  macropinacoid  draw  a  line  parallel  to  the 
b  axis  through  the  end  of  the  a  axis.  For  the 
brachypinacoid  cut  this  line  by  one  parallel  to  the 
a  axis,  drawn  through  the  point  x  on  the  b  axis, 
which  is  cut  by  the  pinacoid.  Through  the  inter- 
section of  these  two  lines  erect  the  vertical  intersec- 
tion of  the  pinacoidal  faces. 

For  the  pyramid  draw  vertical  lines  through  the 
centres  of  the  pinacoid  faces,  then  a  line  parallel 
to  the  front  pyramid  edge,  e,  f,  drawn  through  the 
point  V  where  the  pyramid  cuts  the  vertical  axis, 
will  cut  the  vertical  through  the  centre  of  the 


UNIVERSITY   ] 
&r 


OF 


PLATE  XXXI. 

ORTHORHOMBIC  SYSTEM. 

FIG.  1.  Childrenite.    Axial  Ratio  77 : 1 :  '53 ;  Macropinacoid  {100} ; 
Brachypinacoid  {010} ;  Pyramid  {121}. 

_  FIG.  2.  Calamine.     Axial  Ratio  78 : 1 :  -48  ;  Prism  {110} ;  Pyramid 
{121}  ;  Brachypinacoid  {010}  ;  Brachydome  {031} ;  Macrodome  {301}. 
FIG.  3.  Epsomite.    Axial  Ratio  '99  : 1 :  -57  ;  Prism  {110} ;  Sphenoid 


100         HINTS  FOE  CKYSTAL  DKAWING 

macropinacoid  face  in  g  and  will  be  the  required 
edge  of  the  pyramid. 

A  line  from  the  apex  of  the  pyramid  parallel  to 
one  from  the  apex  of  the  c  axis,  to  cut  the  semi 
b  axis  at  J,  will  cut  the  vertical  through  the  centre 
of  the  brachypinacoid  face  and  will  form  a  short 
edge  of  the  pyramid. 

The  intersections  of  pinacoid  and  pyramid  faces 
will  be,  for  the  macropinacoid  parallel  to  the  side 
pyramid  edges,  for  the  brachypinacoid  parallel  to 
the  front  and  back  pyramid  edges. 

Fig.  2  is  a  somewhat  complicated  crystal  of 
calamine,  axial  ratio  78  : 1  :  "48,  showing  com- 
binations of  unit  prism  {110},  pyramid  {121}.  brachy- 
pinacoid {010}>  brachydome  {031}»macrodome  {301} 
and  basal  pinacoid  {001}- 

The  intersection  of  unit  prism  {110}  with  the 
brachypinacoid  is  found  by  joining  the  ends  of  the 
a  and  b  axes  as  shown,  then  cutting  the  joining  line 
by  a  line  parallel  to  the  a  axis,  through  the  point  g 
on  the  b  axis,  through  which  the  pinacoid  face  {010} 
passes. 

Raise  the  vertical  edge  of  intersection  through 
the  point  of  cutting  /  The  other  edges  may  be 
obtained  by  "parallelism"  and  " repetition  ". 

The  brachydome  replaces  the  prism  and  pina- 
coid at  a  point  p  on  the  c  axis. 

Draw  a  vertical  through  the  centre  of  the  face 
of  the  brachypinacoid,  then  a  line  through  p,  parallel 
to  the  b  axis,  to  cut  the  vertical  gives  the  point  h 
through  which  the  intersection  edge  of  dome  with 
pinacoid  may  be  drawn  parallel  to  the  a  axis. 

The  inclination  of  the  brachydome  may  now  be 


OBTHOKHOMBIC  SYSTEM  101 

obtained  by  a  line  from  h  parallel  to  one  from  the 
apex  of  the  c  axis  to  the  ^  point  g,  of  the  b  axis. 

The  macrodome  passes  through  a  point  at  t,  we 
can  get  its  inclination  and  the  point  where  it  meets 
that  vertical  prism  edge,  which  passes  through  the 
4-  end  of  the  a  axis  by  aid  of  a  line  drawn  through  t, 
parallel  to  one  from  the  -h  end  of  the  c  axis  to  the  ^ 
point  of  the  a,  according  to  the  symbol  {301}.  This 
line  will  cut  the  prism  edge  in  point  X. 

The  domes  give  way  to  the  basal  plane  at  a 
point  on  the  c  axis  marked  i.  Through  i  lines 
parallel  to  the  a  and  b  axes  will  meet  the  lines  of 
inclination  of  the  two  domes  in  points  SS',  through 
which  their  intersections  with  the  base  may  be 
drawn ;  that  from  the  brachydome  parallel  to  the 
brachy  axis ;  that  for  the  macrodome  to  the  macro 
axis. 

We  need  the  intersection  edge  between  the  two 
domes ;  we  can  obtain  it  by  drawing  lines  through 
h  and  t  parallel  respectively  to  the  a  and  b  axes  ; 
where  they  meet  will  be  one  point  /  in  the  intersec- 
tion, the  other  will  of  course  be  where  the  inter- 
sections with  the  base  meet. 

One  other  form  is  present,  the  unit  pyramid 
which  forms  the  lower  end  of  the  crystal.  The 
symbol  is  {121}  and  furnishes  the  clue  for  the  con- 
struction. The  pyramid  cuts  the  -  c  axis  in  point  kt 
a  line  from  k  parallel  to  one  from  +  end  of  a  axis  to 
-  end  of  fi  will  give  the  front  edge  of  pyramid. 
Lines  from  k  on  either  side  parallel  to  one  joining 
the  -  end  of  the  c  axis  to  the  \  points  of  the  +  and 
—  semi  b  will  give  the  side  edges.  The  right  side 
edge  meets  the  pinacoid  face  at  d,  where  it  cuts  a 


102          HINTS  FOE  CKYSTAL  DBA  WING 

vertical  through  the  centre  of  that  face.  A  line 
from  d  parallel  to  the  front  edge  gives  the  inter- 
section of  the  two  forms.  This  intersection  edge 
meets  the  prism  edge  at  e,  from  which  point  the 
intersection  of  pyramid  with  prism  can  be  drawn,  to 
where  the  two  front  edges  meet. 

Fig.  3  represents  a  crystal  of  epsomite,  axial 
ratio  *99  : 1  :  *57.  It  is  a  combination  of  unit  prism 
{110}  and  sphenoid  {111}. 

First  draw  prism  vertical  edges. 

To  obtain  the  intersections  of  prism  and  sphenoid 
draw  verticals  through  the  centres  of  the  prism 
faces  {110}  and  UlO}-  Then  through  apex  of 
crystal  g,  which  is  twice  the  semi  height  of  the  c 
axis  from  the  centre,  draw  a  line  g,  i,  parallel  to  one 
joining  the  +  ends  of  the  //axes.  This  line  will  be 
an  edge  of  the  sphenoid  ;  it  will  cut  the  vertical 
through  the  centre  of  the  prism  face  {110}-  Next 
draw  line  g,  h,  downwards  parallel  to  one  joining  the 
apex  of  the  c  axis  and  point  n,  n  being  the  centre  of 
the  line  joining  the  +  ends  of  the  a  and  b  axes,#,  h, 
will  cut  the  vertical  through  n  in  a  point  h  through 
which  may  be  drawn  an  intersection  edge  of  prism 
and  sphenoid  faces,  parallel  to  the  edge  through  g  ; 
where  this  line  meets  the  vertical  edge  of  prism, 
will  give  a  point  k.  Join  k  to  i  for  intersection  of 
prism  face  {110}  with  sphenoid. 

Then  the  other  upper  face  of  sphenoid  can  be 
got  by  "  parallelism  ". 

The  lower  faces  of  the  sphenoid  will  be  found 
by  drawing  a  line  through  the  lowest  point  of  crystal 
parallel  to  a  line  joining  +  end  of  a  and  —  end  of  b 
axes  or  to  /,  n,  the  line  bisecting  the  angle  between 


OKTHOKHOMBIC  SYSTEM  103 

the  +  ends  of  the  a  and  b  axes.  Let  this  line  parallel 
to  /,  n,  cut  the  vertical  through  the  centre  of  the 
prism  face  {110}-  It  will  be  the  lowest  edge  of  the 
sphenoid.  Draw  through  the  point  p  where  it  cuts 
the  vertical,  a  line  parallel  to  #,  y ;  x,  y  being  drawn 
from  -  end  of  c  axis  to  centre  of  line  joining  +  end 
of  a  axis  to  —  end  of  b.  The  line  drawn  through  p 
will  be  the  intersection  of  sphenoid  face  {111}  and 
prism  face  {110}-  Draw  a  line  parallel  to  x,  y 
through  the  lowest  point  of  crystal,  r,  it  will  cut  the 
vertical  through  the  centre  of  the  prism  face  {110}, 
through  the  point  of  cutting  will  lie  an  intersecting 
edge  parallel  to  /,  n.  The  crystal  is  easily  finished 
by  the  usual  methods. 

Plate  XXXII.  shows  a  crystal  of  chrysolite, 
axial  ratio  *46  : 1  :  '58.  It  is  a  combination  of  the 
unit  pyramid  {111},  unit  prism  {110},  macropina- 
coid  {100}>  brachypinacoid  {010},  macrodome  {101}, 
brachydome  {021}  and  basal  pinacoid  {001} 

The  construction  is  shown  somewhat  fully. 

After  the  axes  have  been  found,  the  lines  join- 
ing the  H  axes  are  put  in,  then  from  the  0  plan 
the  points  p,  q,  r,  the  points  of  intersection  of  the 
prism  and  pinacoids  are  dropped  by  verticals  to  the  C 
projection  in  p\  q\  r',  and  through  p ',  </,  the  vertical 
intersection  edges  are  raised. 

The  prism  and  pinacoids  are  replaced  at  a  point 
0  on  the  vertical  axis  by  the  domes  and  unit 
pyramid. 

Through  0  draw  lines  parallel  to  the  a  and  b 
axes,  also  lines  parallel  to  those  joining  the  axes. 
On  the  latter  will  lie  the  intersections  of  unit  prism 
and  pyramid,  they  will  meet  the  vertical  intersections 


104          HINTS  FOE  CKYSTAL  DKAWING 

of  the  pinacoids  and  prism.  Through  the  points 
of  meeting  e,  fy  the  intersections  of  pinacoids  and 
domes  may  be  drawn,  parallel  respectively,  to  the 
macro  and  brachy  axes. 

We  may  next  find  the  inclination  lines  of  the 
domes.  Make  OA  equal  the  vertical  axial  height, 
join  A  to  g,  then  draw  a  line  parallel  to  Ag,  through 
point  h ;  this  line  will  be  the  inclination  line  of  the 
macrodome  {101}- 

For  the  inclination  of  the  brachydome  {021}  join 
A  to  k,  Ok  being  equal  to  |  the  semi  b  axis ;  from 
the  centre  of  the  intersection  of  brachypinacoid  and 
dome  i  draw  a  line  parallel  to  Ak.  This  is  the 
line  of  inclination  of  the  brachydome. 

Between  the  faces  of  the  macrodome  {101}  and 
brachydome  {021}  is  a  face  of  the  unit  pyramid 
{111}-  Its  intersection  with  the  unit  prism  {110} 
will  evidently  be  e,  f. 

To  find  its  intersection  with  the  macrodome 
draw  the  pyramid  edge  A  B,  then  through  h',  the 
point  where  the  inclination  line  of  the  macrodome 
cuts  the  c  axis,  a  line  parallel  to  the  b  axis,  to  cut 
the  edge  A  B  ;  join  the  point  of  cutting  d  to  e  for  the 
intersection  of  the  forms. 

Next  find  the  intersection  edge  between  the 
pyramid  {111}  and  the  brachydome.  This  is  done 
in  a  similar  way,  by  joining/ to /',  the  point  where 
the  pyramid  edge  and  line  of  inclination  of  the 
brachydome  meet. 

The  basal  pinacoid  {001}  cuts  the  vertical  axis 
produced  in  X.  Through  X  draw  lines  parallel  to 
the  a  and  b  axes  to  cut  the  line  of  inclination  of  the 
macrodome  and  the  unit  pyramid  edge  AB,  in  points 


PLATE  XXXII. 

ORTHORHOMBIC  SYSTEM. 

Chrysolite.  Axial  Ratio  '46  : 1 :  '58  ;  Macropinacoid  {100} ;  Macro- 
dome  {101}  ;  Brachypinacoid  {010}  ;  Brachydome  {021} ;  Unit  Prism 
{110} ;  Unit  Pyramid  {111} ;  Basal  Pinacoid  {001}. 


106          HINTS  FOE  CKYSTAL  DKAWING 

n,  n'.  Through  these  points  the  intersections  of  base 
with  dome  and  pyramid,  may  be  drawn  parallel  to 
the  intersections  of  those  forms  with  prism  and 
pinacoids. 

Complete  the  figure  by  the  usual  methods. 

Exercises — Problems  for  solution. 
Orthorhombic  system— 

Andalusite,  axial  ratio  '98  : 1 :  7. 

{1101;  {001{;  {101}. 
Staurolite,  axial  ratio  '47  : 1  :  '68. 

{110};  {010};  {001};  {101K 
Olivine,  axial  ratio  "47  : 1  : 59. 

{110};  {120};  {010};  {021};  {101};  {111}- 
Cordierite,  axial  ratio  '59  : 1  : 56. 

{110};  {130};  {010};  {001}- 


CHAPTER  XIII. 

MONOCLINIC     SYSTEM— CONSTRUCTION     OF    AXES— SIMPLE 
FORMS— COMBINATIONS. 

PLATES  XXXIII.  TO  XXXVI. 

THE  axes  of  the  monoclinic  system  are  all  unequal ; 
they  are  lettered  as  in  the  orthorhombic  system, 
a,  b,  c. 

The  a  axis  is  at  right  angles  to  the  b.  The  b  is  at 
right  angles  to  the  c.  The  a  is  inclined  to  the  c. 

The  angle  the  a  axis  makes  with  the  c  is  measured 
from  +  c  to  —  a,  it  is  represented  by  the  Greek  letter  |3. 

The  a  axis  is  sometimes  called  the  clinodiagonal 
axis,  the  b  axis  the  orthodiagonal. 

Plate  XXXIII.  Fig.  1  shows  the  construction  of 
the  monoclinic  axes.  The  assumed  axial  ratio  is 
•65  :  1  :  '55.  The  fi  angle  64°. 

Begin  with  the  0  plan  in  exactly  the  same  way 
as  for  the  isometric  and  tetragonal  systems.  Make 
the  b  axis  equal  1,  also  make  the  line  of  direction  of 
the  a  axis  equal  1 ;  divide  it  into  tenths  and  mark 
the  point  *65  at  Y. 

Thus  far  the  construction  differs  only  from  the 
isometric  and  tetragonal  in  axial  ratio,  and  if  axis  a 
were  at  right  angles  to  both  b  and  c,  XY  would 
represent  it  in  0  plan.  But  axis  a  is  inclined  to  c  at 
an  angle  /3  upwards  and  backwards.  Looked  at 
sideways,  in  profile,  its  position  with  regard  to  c  is  as 

107 


108         HINTS  FOE  CEYSTAL  DEAWING 

shown  in  Fig.  2.  If  the  a  axis  were  at  right  angles  to 
the  c  its  position  seen  in  profile  would  be  horizontal, 
it  would  be  coincident  with  YX' ;  but  it  is  inclined, 
it  has  taken  the  position  of  the  line  from  +  a!  to  —  a', 
it  makes  the  angle  /3  with  the  vertical  axis.  Now 
point  -f  a!  lies  on  an  arc  drawn  with  the  radius  X'  Y ', 
and  if  we  draw  a  vertical  through  it  to  cut  the  line 
YX,  we  find  it  lies  under  a  point  pf  on  that  line ; 
the  distance  p'X'  equals  the  distance  point  +  a!  is 
from  the  vertical  axis,  is  in  fact  equal  to  the  plan  of 
the  semi  a  axis. 

It  is  this  distance  from  p'  to  X'  that  we  have  to 
mark  off  on  X  Y  the  line  of  direction  of  the  a  axis  on 
the  0  plan.  It  gives  us  point  p,  it  is  from  point  p 
the  length  of  the  plan,  not  from  point  Y  the  actual 
length,  that  we  are  to  drop  the  vertical  projector  to 
the  C  projection  ;  the  ^  distance  we  have  to  mark  on 
it  for  the  apparent  depression  is  half  e  X ;  thus  we 
get  point  S. 

But  even  in  S  we  have  not  obtained  the  position 
for  the  4-  end  of  the  a  axis  in  C  projection,  for  point 
S  would  be  the  position  of  point  p  supposing  X,  p 
were  merely  a  horizontal  line,  X,  p  being  the  true 
length ;  as  we  have  seen  X,  p  is  the  plan  of  an 
inclined  line,  the  end  of  which  is  vertically  below  p  ; 
to  find  how  far  below  we  again  refer  to  the  profile, 
Fig.  2,  p,  4-  a',  gives  us  the  distance  below. 

In  C  projection  we  know  that  all  vertical 
distances  retain  their  true  length,  therefore  we  take 
the  distance  //,  -f  a'  and  mark  it  directly  on  the  pro- 
jector through  S  downwards  ;  it  gives  us  the  position 
for  the  +  end  of  the  a  axis  ;  producing  the  axis  beyond 
the  centre  we  mark  off  an  equal  distance  for  —  a. 


{001}. 


PLATE  XXXIII. 
MONOCLINIC  SYSTEM. 

FIG.  1.  Construction  of  Axes,  assumed  Axial  Ratio  '65  : 1 :  P55. 
FIG.  2.  Profile  of  a  and  c  Axes. 
FIG.  3.  Unit  Prism  {110} ;  Base  {001}. 
FIG.  4.  Orthopinacoid  {100} ;  Clinopinacoid  {010} ;  Basal  Pinacoid 


110          HINTS  FOE  CKYSTAL.  DKAWING 

The  b  axis  being  obtained  in  exactly  the  same 
way  as  for  the  other  systems  needs  no  explanation. 

The  problem  of  the  inclined  axis  might  be  solved 
by  trigonometry,  but  we  are  pledged  not  to  appeal 
to  trigonometry  and  the  profile  method  requires  less 
mental  effort. 

Fig.  3  shows  the  combination  of  unit  prism  {110} 
with  the  base  {001}-  It  should  offer  no  difficulty 
when  once  the  axis  construction  is  mastered. 

Fig.  4  is  the  combination  of  orthopinacoid  {100} 
clinopinacoid  {010}  and  basal  pinacoid  {001};  the 
construction  is  perfectly  evident. 

Plate  XXXIV.  Figs.  1  and  2  are  the  pyramids 
{111}  and  {111}  each  consisting  of  only  four  faces. 
Fig.  3  is  a  combination  of  the  two  pyramids. 

Fig.  4  is  a  crystal  of  gypsum,  axial  ratio  '7:1:  *4, 
angle  /3  =  80°  42'.  It  is  composed  of  unit  pyramid 
{!!!}>  unit  prism  {110}  and  clinopinacoid  {010}- 
The  latter  cuts  the  b  axis  of  the  unit  prism  at  point  e. 
Join  the  ends  of  the  a  and  b  axes  and  cut  the  line 
joining  them  by  one  through  point  e  parallel  to  the 
a  axis,  raise  the  vertical  edge  of  the  pinacoid  and 
prism  through  the  meeting  point  of  these  lines. 

The  pyramid  edge  meets  the  prism  edge  at 
point  X.  From  point  X  draw  the  unit  pyramid  edge 
to  cut  the  c  axis  at  the  proper  inclination. 

Because  the  unit  pyramid  of  this  system  is  com- 
posed of  only  four  faces,  the  front  edge  must  be 
continued  beyond  and  behind  the  V  axis  until  it 
meets  the  edge  of  the  unit  prism  through  the  —  a 
axis. 

From  the  point  of  meeting  of  pyramid  and 
prism  edges,  X}  the  intersection  edges  of  the  two 


£    S 

O     £L 


112         HINTS  FOE  CEYSTAL  DEAWING 

forms  may  be  drawn  ;  they  will  be  lines  parallel  to 
the  front  edges  of  the  unit  pyramid  X,  0,  and  X,  n. 
The  intersection  edges  of  pyramid  and  pinacoid  will 
be  parallel  to  the  centre  edge  of  the  pyramid ;  they 
will  cut  the  back  vertical  intersections  of  prism  and 
pinacoid  in  p  and  q.  Join  p  and  q  to  r  for  further 
intersection  edges  of  pyramid  and  prism. 

The  crystal  may  now  be  completed  by  the  usual 
methods. 

Fig.  5  is  a  crystal  of  sphene,  axial  ratio  75  : 1 : 
*85,  /?  angle  60°  17'.  It  is  a  combination  of  unit 
pyramid  {111}  and  unit  prism  {110}. 

The  pyramid  replaces  the  prism  at  points  p,  p' 
on  the  c  axis. 

Draw  the  vertical  prism  edges  e,  I  and  e\  k  through 
the  ends  of  the  a  axis  and/,  h',  g,  i  through  the  ends 
of  the  b,  cut  them  by  lines  through  p  and  p'  parallel 
to  the  a  and  b  axes  in  points  e,  e',  f,  g,  h,  i.  Through 
the  points  e,  d  draw  the  central  edges  of  the  pyramid 
e,  k,  e',  I  parallel  to  lines  joining  the  4-  end  of  the  a  to 
the  -f  end  of  the  c  axis  and  the  -  end  of  the  a  to 
the  —  end  of  the  c  axis  :  take  them  through  the 
vertical  axis  producing  them  to  cut  the  prism  edges 
through  the  +  and  —  ends  of  the  a  axis. 

To  complete  the  figure  it  is  only  necessary  to 
join  the  points/,  g,  h,  i  to  the  points  k,  I  where  the 
central  edges  of  prism  and  pyramid  meet. 

Plate  XXXV.  gives  combinations  of  other 
monoclinic  forms. 

The  clinodomes  shown  in  Fig.  1,  the  orthodomes 
in  Fig.  2,  need  no  direction ;  the  construction  is 
apparent. 

Fig.  3  is  a  crystal  of  amphibole,  axial  ratio  *55  : 


PLATE  XXXV. 

MONOCLINIC  SYSTEM. 

FIG.  1.  Clinodomes  {Oil}. 

FIG.  2.  Orthodomes  {101}. 

FIG.  3.  Amphibole.  Axial  Ratio  -55  : 1 :  -29  ;  0  73°  58'.  Unit  Prism 
{110} ;  Clinopinacoid  {010} ;  Clinodome  {Oil}. 

FIG.  4.  Epidote.  Axial  Ratio  1-58  : 1 :  T8 ;  0  64°  37'.  Orthopinacoid 
{100} ;  Orthodome  {110} ;  Basal  Pinacoid  {001} ;  Unit  Pyramids  {111} 
{111}  {111}  {111}. 


8 


114         HINTS  FOE  CBYSTAL  DBA  WING 

1 :  '29,  ft  angle  =  73°  58'.  It  is  a  combination  of 
unit  prism  {110},  clinodome  {011}  and  clinopinacoid 
{010}. 

After  finding  the  axes  and  joining  the  ends  of 
the  a  and  b,  a  line  parallel  to  the  a  axis  through 
point  p,  where  the  pinacoid  cuts  the  b  axis,  will  cut 
the  lines  joining  the  ends  of  the  axes  in  points, 
through  which  may  be  raised  the  vertical  intersec- 
tions of  prism  and  pinacoid. 

The  inclination  of  the  clinodome  which  passes 
through  a  point  i  on  the  vertical  axis  produced  will 
be  a  line  it  n  parallel  to  one  joining  the  +  ends  of  the 
b  and  c  axes,  it  will  cut  a  vertical  through  the 
centre  of  the  pinacoid  face,  so  giving  a  point  n 
through  which  the  intersection  with  that  face  may 
be  drawn  parallel  to  the  a  axis.  The  top  edge  of 
the  dome  is  drawn  through  point  i  also  parallel  to 
the  a  axis,  it  cuts  the  prism  edges  in  k  and  /.  k  and 
/  joined  to  points  e,f,  g,  h,  will  give  the  intersec- 
tions of  domal  with  prismatic  faces. 

The  base  of  the  crystal  can  be  finished  by 
"  parallelism  ". 

Fig.  4  is  a  crystal  of  epidote,  axial  ratio  1*58: 
1:1-8,  ft  angle  =  64°  37'.  It  is  acombination  of 
orthopinacoid  U00}>  orthodorne  {110}.  basal  pina- 
coid {001},  unit  pyramids  {Til!  {Ill}  {111}  and 
{111}. 

It  is  prismatic  in  the  direction  of  the  ortho 
axis.  The  intersection  of  orthopinacoid  and  basal 
pinacoid  may  be  obtained  by  verticals  through  the 
ends  of  the  a  axis,  cut  by  lines  parallel  to  the  a  axis 
drawn  through  point  x  the  point  where  the  base 
intersects  the  c  axis. 


MONOCLINIC  SYSTEM  115 

The  dome  cuts  the  orthopinacoid  face  in  point  r, 
through  which  may  be  drawn  their  basal  intersec- 
tion. The  inclination  of  the  dome  will  be  a  line 
parallel  to  the  -  front  edge  of  the  unit  pyramid. 

To  find  the  intersections  of  the  various  forms 
with  the  unit  pyramid,  it  will  be  useful  to  have  the 
edges  of  the  pyramid  shown,  because  all  the  inter- 
sections will  be  parallel  to  one  or  other  of  these 
thus  :— 

The  intersection  with  the  base  will  be  parallel 
to  the  edges  joining  the  a  and  b  axes. 

That  with  the  orthopinacoid,  will  be  parallel  to 
those  joining  the  b  and  c  axes. 

That  with  the  dome  is  parallel  to  the  front  and 
back  edges.  Thus  the  crystal  may  be  completed. 

Plate  XXXVI.  gives  a  crystal  of  orthoclase, 
axial  ratio  '66  :  1  :  '56  ;  angle  /3  =  63°  57'.  It  is  a 
combination  of  unit  prism  {110},  unit  pyramid  {111}, 
clinopinacoid  {010},  clinoprism  {130},  basal  pinacoid 
{001},  orthodomes  {101}  and  {201}- 

The  front  edge  of  the  unit  prism  is  drawn  verti- 
cally through  the  +  end  of  the  a  axis.  The  +  ends  of 
the  a  and  b  axes  are  joined.  The  line  joining  them 
will  be  the  trace  of  the  unit  prism  on  the  axial  plane 
which  includes  the  a  and  b  axes.  This  plane  is 
parallel  to  the  basal  plane  {001};  we  may  there- 
fore call  the  intersections  of  prisms  and  pinacoids 
with  it,  their  basal  traces. 

We  have  the  front  and  back  edges  of  the  unit 
prism  through  the  4-  and  —  ends  of  the  a  axis.  At 
a  point  U  on  its  basal  trace  the  prism  {110}  meets 
the  prism  {130}-  We  can  draw  the  basal  trace  of 

this  latter  prism  from  U.     It  will  be  a  line  parallel  to 

8* 


116          HINTS  FOE  CKYSTAL  DEAWING 

one  drawn  from  the  -f  end  of  the  a  axis  to  the  J  point 
on  the  b  axis.  It  will  meet  the  basal  trace  of  the 
clinopinacoid  {010}.  a  line  VW  drawn  through  point 
2,  the  point  where  the  pinacoid  cuts  the  b  axis, 
parallel  to  the  a  axis ;  the  point  of  meeting  is 
lettered  V.  Through  U  and  V  we  can  raise  the 
vertical  intersections  of  prisms  and  pinacoid. 

The  front  edge  of  the  unit  prism  is  cut  by  the 
base  in  point  T.  From  T  draw  a  line  parallel  to 
the  a  axis  to  cut  the  vertical  axis  in  S ;  this  will  be 
the  inclination  line  of  the  basal  plane. 

The  intersection  of  basal  plane  and  unit  prism 
will  be  a  line  from  T  parallel  to  the  basal  trace  of 
the  prism.  It  meets  the  intersection  edge  of  unit 
and  clinoprisms  in  point  R. 

The  intersection  of  basal  plane  and  clinoprism 
will  be  a  line  from  R  parallel  to  X  Y,  the  basal  trace 
of  the  prism.  It  will  meet  the  intersection  of  clino- 
prism and  pinacoid  in  point  P. 

The  intersection  of  basal  plane  with  clinopina- 
coid will  be  a  line  from  P  parallel  to  the  a  axis. 

We  have  next  to  find  the  position  of  the  face 
of  the  orthodome  {201}  and  its  intersection  edges. 
Its  lower  front  face  {201}  cuts  the  front  edge  of 
the  unit  prism  in  point  0.  Its  inclination  line  is 
drawn  from  O  parallel  to  a  line  joining  the  %  point 
of  the  a  axis  to  the  -  end  of  the  c  axis. 

The  intersection  of  the  orthodome  face  {201} 
with  the  unit  prism  is  got  by  drawing  through  the 
point  e  where  its  line  of  inclination  cuts  the  a  axis  a 
line  e,f,  parallel  to  the  b  axis,  to  cut  the  basal  trace 
of  the  prism  inf.  Through  the  point  of  cutting  the 
required  intersection  ON  can  be  drawn. 


118          HINTS  FOE  CKYSTAL  DRAWING 

The  orthodome  face  {201}  gives  place  to  the 
ortho-dome  face  {101}  at  point  L.  The  intersection 
of  the  two  orthodomes  will  be  parallel  to  the  b  axis. 

Draw  the  line  of  inclination  of  the  orthodome 
{101!-  It  joins  the  +  end  of  the  a  and  -  end  of  the 
c  axis,  then  through  the  -  end  of  the  c  axis  its  inter- 
section with  the  basal  plane. 

One  other  form  is  present,  the  unit  pyramid 
{111}-  Its  lower  right-hand  face  {111}  cuts  the 
clinopinacoid  at  a  point  M.  Its  intersection  with 
the  pinacoid  is  of  course  parallel  to  its  own  front 
edge.  Its  intersection  with  the  face  of  the  prism 
{130}  is  found  by  drawing  a  vertical  through  the  ^ 
point  of  the  semi  b  axis  to  cut  the  line  joining  the  + 
end  of  the  b  to  the  —  end  of  the  c  axis  in  p  ;  join  p, 
the  point  of  cutting  to  the  +  end  of  the  a  axis,  and 
make  the  edge  of  intersection  parallel  to  it  Nnf. 

The  intersection  of  the  unit  pyramid  and  the 
orthodome  {201}  will  be  parallel  to  a  line  joining 
the  point  where  the  inclination  line  of  the  orthodome 
meets  the  front  edge  of  the  pyramid  point  L,  to  where 
line  e,  f  cuts  the  line  from  -f  end  of  a  to  -f-  end  of  b 
axis. 

The  intersection  of  the  pyramid  with  the  ortho- 
dome  {101}  will  be  parallel  to  the  front  edge  of  the 
pyramid. 

The  intersection  with  the  basal  plane  is  MK 
parallel  to  a  line  joining  the  -f  a  and  +  b  axes. 

The  rest  is  "  parallelism  "  and  "  repetition  ". 

Exercises — Problems  for  solution. 
Monoclinic  system- 
Gypsum,  axial  ratio  '69  : 1  :  '4  ;  ft  80°  42' 

{010};  {110};  {111}. 


MONOCLINIC  SYSTEM  119 

Hornblende,  axial  ratio  -55  : 1  :  '29  ;  ft  73°  58'. 

{110};  {010!;  {OIK;  ilOl}- 
Epidote,  axial  ratio  1-58:1:1-80;  ft  64°  37'. 

{100} ;  1001};  {101};  {110}- 

Orthoclase,  axial  ratio  -66  : 1  :  '56  ;  ft  63°  57'. 
{010! ;  {110} ;  {001} ;  {101} ;  {201} ;  {011} ;  {lilt- 
Orthoclase,  {001};  {010};  {110};  {201}- 


CHAPTEE  XIV. 

TRICLINIC  SYSTEM— CONSTRUCTION  OF  AXES— SIMPLE 
FORMS— CONBINATIONS. 


PLATES  XXXVII.  TO  XXXIX. 

IN  the  triclinic  system  there  are  three  axes,  all  un- 
equal and  intersecting  at  oblique  angles.  They  are 
lettered  a,  b,  c.  The  angle  between  the  c  or  vertical 
axis  and  the  a  axis  is  measured  from  the  +  end  of 
one  to  the  +  end  of  the  other,  it  is  lettered  ft.  The 
angle  between  the  c  and  b  axes  is  measured  from 
the  +  end  of  one  to  the  4-  end  of  the  other ;  it  is 
lettered  a.  The  angle  between  the  a  and  b  axes  is 
measured  from  the  +  end  of  the  one  to  the  +  end 
of  the  other  and  is  lettered  y. 

Plate  XXXVII.  shows  the  construction  of  the 
triclinic  axes.  It  is  dependent  on  the  0  plan  and 
profiles  of  the  axes. 

The  a  axis  is  in  the  same  direction  as  in  the 
isometric  system  but  has  this  difference,  the  +  end 
in  the  triclinic  system  is  either  depressed  or  ele- 
vated a  certain  number  of  degrees,  so  that  the 
axis  is  no  longer  at  right  angles  to  the  c  or  vertical 
axis,  its  position  is  similar  to  the  a  axis  of  the 
monoclinic  system. 

The  b  axis  is  neither  at  right  angles  to  the  c  nor 

120 


TEICLINIC  SYSTEM  121 

to  the  a  axis,  therefore  its  position  is  quite  altered 
from  that  of  the  isometric  a2  axis. 

We  will  assume  the  lengths  given  on  the  plate 
for  the  three  axes,  making  as  usual  the  b  axis  = 
unity.  The  angles  are  also  assumed  f$  =  140°,  a  = 
75°,  y  ---  135°. 

We  start  by  finding  the  direction  of  the  a  axis 
in  the  usual  manner,  then  find  its  plan  by  a  profile 
(see  Fig.  2)  exactly  as  for  the  monoclinic  a  axis. 
Having  thus  obtained  the  plan  of  its  semi-length,  we 
mark  it  on  the  0  plan  as  shown  at  a,  x. 

We  next  find  the  plan  of  the  semi  b  axis,  in  the 
same  way  by  a  profile,  shown  at  Fig.  3. 

Having  then  the  plans  of  the  semi  a  and  b  axes, 
it  would  at  first  appear  quite  easy  to  construct  the 
0  plan  by  making  the  plan  of  the  b  axis  at  an  angle 
of  105°  with  the  plan  of  the  a  axis. 

This,  however,  would  not  be  correct,  the  plan  of 
the  y  angle  is  not  the  true  angle  105°,  and  it  is  the 
plan  we  require. 

To  get  it  we  must  make  use  of  yet  another  pro- 
file, giving  another  plan.  This  plan  is  of  the  line 
that  would  join  the  +  ends  of  the  a  and  b  axes. 

With  a  little  thought  we  can  attain  this  plan. 
Make  a  triangle  (see  Fig.  4)  having  the  semi  a  and 
b  axes  for  sides  at  an  angle  to  each  other  equal  the 
true  y  angle  105°,  the  other  side  of  the  triangle, 
the  hypotenuse,  will  be  the  true  length  of  the  line 
joining  the  axes. 

We  have  the  true  length,  but  it  is  the  plan  of 
this  line  we  require  to  complete  our  O  plan. 

To  obtain  the  plan  we  need  to  know  the  height 
the  end  of  the  b  axis  is  above  the  level  of  the  end 


122          HINTS  FOE  CEYSTAL  DEAWING 

of  the  a  axis.  The  +  end  of  the  a  axis  is  the  dis- 
tance r,  t  below  the  centre  of  the  vertical  axis  ;  the 
+  end  of  the  b  axis  is  the  height  u,  s  above  the  centre. 
The  height  then  of  the  end  of  the  b  axis  above  the 
end  of  the  a  is  us  +  rt. 

Now  we  have  all  the  factors  necessary  for 
finding  the  plan  of  the  line  joining  the  ends  of  the 
a  and  b  axes.  We  find  it  thus  :— 

Draw  a  horizontal  line  hhf  (see  Fig.  5)  at  h, 
raise  a  perpendicular  hg  =  us  +  rt,  from  g  with 
radius  the  true  length  of  the  line  taken  from  Fig.  4 
cut  hh!  in  i.  Join  i  to  g,  then  h,  i  will  be  the  plan  of 

ff,  i- 

We  can  now  finish  our  0  plan  of  the  axes. 
We  have  already  found  the  plan  of  the  semi  a  axis, 
from  the  end  x  with  radius  =  plan  of  semi  b  axis 
make  an  arc,  from  the  other  end  a  with  radius  = 
plan  of  line  joining  ends  of  a  and  b  axes,  cut  the  arc 
in  e,  the  0  plan  is  now  complete. 

We  drop  the  ends  of  the  axes  by  vertical  pro- 
jectors to  the  C  projection  as  usual,  find  points  r 
and  u  which  would  be  the  ends  of  the  plans  of  the 
axes  if  they  were  horizontal ;  to  get  the  actual  posi- 
tions mark  from  r  downwards  the  distance  t,  r  taken 
from  the  profile  Fig.  2  of  the  a  axis  and  for  the  b 
axis  mark  upwards  from  u  the  height  u,  s  taken  from 
Fig.  3  the  profile  of  the  b  axis. 

Plate  XXXVIII.  shows  the  simple  forms  of  the 
triclinic  system,  each  composed  of  only  two  faces. 

Figs.  1,  2,  3,  4  are  the  four  pyramids  {111} 
{111}  {111}  {111}. 

Figs.  5  and  6  are  the  prisms  {110}     {110}- 

Figs.  7  and  8  are  the  pinacoids  {100}  and  {010}, 


axis  bsl 


PlcLtt  of  list-e.  joLtUtig  CL  tvul  b  <UGet 

' 


PLATE  XXXVII. 

TBICLWIC  SYSTEM. 
Construction  of  Axes. 


124          HINTS  FOE  CRYSTAL  DRAWING 

Figs.  9  and  10  are  macrodomes  {101}  and  {101} 

Figs.  11  and  12  are  the  brachy domes  {Oil! 
and  {Oil}- 

These  with  the  basal  pinacoid  {001}  combine  to 
form  the  crystals  of  the  system. 

Plate  XXXIX.  shows  a  crystal  of  axinite,  axial 
ratio  -49  :  1  :  '48.  Angles  a  =  82°  54',  ft  =  91°  52', 
y  131°  32'.  It  is  a  combination  of  the  forms, 
brachyprism  {010},  macropinacoid  {100}.  two 
prisms  {110}  and  UIO}>  two  pyramids  {111}  and 
{III}  and  the  macrodome  {201}- 

The  construction  for  finding  the  axes  is  shown, 
though  for  description  the  student  is  referred  to 
p.  121. 

Having  constructed  the  axes,  join  the  ends  of 
the  a  and  b  axes  of  the  0  plan. 

The  pinacoids  cut  the  axes  in  points  PP'  shown 
on  the  0  plan.  Draw  a  line  through  P  parallel  to 
the  b  axis  and  one  through  P'  parallel  to  the  a, 
where  these  lines  cut  those  joining  the  axes  will 
give  points  n,  nf,  g,  gf,  drop  these  points  by  vertical 
projectors  to  the  C  projection,  on  these  verticals 
will  be  the  vertical  intersections  of  the  prisms  and 
pinacoids  and  by  "  repetition  "  we  can  find  the  face  of 
the  prism  {010}  and  macropinacoid  {TOO}- 

At  the  point  //on  the  c  axis  (C projection)  the 
prisms  and  pinacoids  give  place  to  pyramids  and 
dome. 

Through  point  H  draw  lines  parallel  to  the  a  and 
b  axes  ff,  ee',  and  join  the  ends.  The  intersection 
edge  between  the  pyramid  {111}  and  the  prism 
{1TO}  will  lie  on  the  line  from  e  to/ 

We  will  next  follow  carefully  the  construction 


126         HINTS  FOE  CKYSTAL  DBA  WING 

for  the  several  intersection  edges  of  the  pyramid 
face  {111}  with  the  other  forms. 

We  have  its  intersection  with  prism  {110}»  line 
e,  f.  It  is  cut  off  by  the  vertical  intersections  of 
prism  and  pinacoid  in  point  g'  and  n'.  Draw  a  line 
f rom  e  parallel  to  one  from  the  +  end  of  the  a  axis 
to  the  +  end  of  the  c,  carry  it  through  and  beyond 
the  c  axis,  until  it  meets  a  vertical  raised  through 
the  centre  of  the  further  macropinacoid  face  {100}, 
through  the  point  of  meeting  we  can  draw  the 
intersection  of  pyramid  and  pinacoid,  it  will  be 
parallel  to  a  line  joining  the  —  end  of  the  b  to  the 
+  end  of  the  c  axis,  it  meets  the  vertical  edges  of 
the  pinacoid  in  h  and  o. 

The  intersection  of  pyramid  and  brachypinacoid 
{010}  will  be  line  g,  g'  drawn  parallel  to  e,  i  the  line  of 
inclination  of  the  pyramid. 

Carry  the  line  of  inclination  to  cut  a  vertical 
through  the  —  end  of  the  a  axis  in  point  i.  A  line 
joining  the  point  of  cutting  i  to /will  be  the  inter- 
section edge  of  pyramid  and  prism  face  {110}  and 
will  join  point  h  to  g. 

It  may  be  well  to  test  the  accuracy  of  point  i, 
since  the  intersection  is  somewhat  acute ;  it  may  be 
done  thus : — 

Draw  on  the  0  plan  at  F  a  line  perpendicular 
to  the  a  axis,  YZ,  make  it  equal  the  semi  c  axis,  join 
Z  to  V.  YVZ  is  a  profile  of  line  e,  i  cut  through  the 
a  and  c  axes.  The  distance  YZ  should  equal/,  i  of 
the  C  projection  if  correct. 

The  pyramid  face  {111}  is  cut  by  the  macrodome 
{201}- 

The  macrodome  meets  the  macropinacoid  {100} 


f 


(1) 


PLATE  XXXIX. 

TRICLINIC  SYSTEM. 

FIG.  1.  Axinite.  Axial  Ratio  '49 : 1 :  '48  ;  a  82°  54' ;  ft  91°  52' ;  y  131° 
32'.  Macropinacoid  {100} ;  Two  Unit  Prisms  {110}  {110} ;  Macrodome  {201} ; 
Two  Unit  Pyramids  {111}  {111} ;  Brachyprism  {010}. 

FIGS.  2,  3,  4.  Construction  for  Axes. 


128          HINTS  FOE  CEYSTAL  DEAWING 

in  a  line  of  intersection  m,  m'  which  is  parallel  to  the 
b  axis.  From  the  centre  of  this  intersection  edge 
draw  the  line  of  inclination  for  the  macrodome, 
{201 }»  it  will  be  a  line  parallel  to  one  from  the  \ 
point  of  the  semi  a  axis  to  the  4-  end  of  the  c. 

Where  this  line  of  inclination  of  the  macrodome 
crosses  the  line  e,  i  will  give  one  point  k  through 
which  to  draw  the  intersection  of  pyramid  and 
macrodome.  Another  will  be  point  d,  found  by 
drawing  through  v,  where  the  macrodome  meets  the 
c  axis  produced,  a  line  parallel  to  the  b  axis,  which 
is  the  trace  of  the  macrodome  on  the  vertical  plane 
containing  the  c  and  b  axes,  at  d  it  meets  the  trace 
of  the  pyramid  on  that  plane,  the  line  /,  d  parallel 
to  one  joining  the  -  end  of  the  b  to  the  +  end  of 
the  c  axis. 

The  intersection  edge  of  pyramid  and  macrodome 
k,  d,  will  meet  that  of  prism  and  pyramid  in  k.  k 
joined  to  m  will  be  the  intersection  of  macrodome 
and  prism.  We  may  check  the  accuracy  of  this  last 
intersection,  by  raising  a  vertical  through /to  cut  the 
line  through  i\  d  in  /;  k',  m  produced  should  pass 
through  /. 

The  intersection  edge  between  macrodome  {201} 
and  prism  {110}  is  obtained,  by  producing  the  line 
0,  d  to  the  right  to  cut  a  vertical  raised  through  the 
+  end  of  the  b  axis  in  point  m!.  Join  m  to  m!  for 
the  intersection  edge  between  macrodome  and 
prism. 

One  other  form  is  present.  The  pyramid  {111} 
cuts  the  prism  in  point  X.  From  X  a  line  parallel  to 
that  joining  the  +  ends  of  the  a  and  b  axes  will  give 
the  intersection  of  pyramid  and  prism. 


TKICLINIC  SYSTEM  129 

The  intersection  of  this  pyramid  with  the  brachy- 
pinacoid  will  be  parallel  to  line  e,  i. 

The  intersection  of  the  pyramid  with  the  dome 
{201}  will  join  X  to  v  the  point  where  both  cut  the 
c  axis  produced. 

The  intersection  of  pyramids  {111}  and  {111} 
will  be  parallel  to  line  e,  i,  it  will  meet  the  short 
intersection  edge  between  the  pyramid  {111}  and 
the  prism  {Il0}>  0,  <?,  which  is  parallel  to  f,  k. 

A  line  from  q  to  r  will  complete  the  crystal. 
This  latter  line  is  the  intersection  edge  between  the 
pyramid  {111}  and  the  prism  {HO}-  It  should  be 
parallel  to  a  line  joining  /'  to  point  if  if  worked 
correctly. 

Exercises — Problems  for  solution. 
Triclinic  system — 

Plagioclase,  axial  ratio  '63  : 1 :  '56  ;  a  94°  3' ; 
ft  116°  29';  y  88°  9'.  {010} ;  {110} ;  {110} ; 
{001};  UOl}- 

Kyanite,  axial  ratio  '9  : 1 :  7  ;  a  90°  5' ;  ft  101° 
2';  y  105° 44'.  {100};  {010};  {001};  {110}; 

{no};  {210}. 


A  CHAPTEE  ON  TWINS. 

TWINNED  CRYSTALS  IN  THE  ISOMETRIC— TETRAGONAL- 
HEXAGONAL—  ORTHOBHOMBIC— MONOCLINIC— AND  TBI- 
CLINIC  SYSTEMS. 

PLATES  XL.  TO  XLIV. 

FOR  the  construction  of  twinned  forms  the  profile 
method  is  most  useful. 

Plate  XL.  shows  a  twinned  octahedron  or  spinel 
twin.  It  is  a  twin  of  the  isometric  system.  The 
twinning  plane  is  parallel  to  an  octahedral  face. 

The  small  figure  3  shows  the  twinning  plane  in 
situ.  The  twinning  axis  is  indicated  by  the  finely 
dotted  line. 

To  draw  the  twinned  crystal,  begin  by  drawing 
an  octahedron,  making  an  0  plan  of  the  H  .axes 
above.  Bisect  the  edges  joining  the  4-  end  of  the 
'0i  axis  to  the  -  end  of  the  02,  the  +  end  of  the  a2  to 
the  —  end  of  the  a^  the  edges  joining  the  4-  ends  of 
the  «!  and  a2  axes  to  the  +  end  of  the  a9  the  edges 
joining  the  —  ends  of  the  a^  and  az  axes  to  the  —  end 
of  the  aB.  The  bisection  of  the  edges  will  give  us 
six  points  through  which  to  draw  the  edges  of  the 
twinning  plane,  shown  shaded  in  the  small  figure  3. 

We  know  that  the  twinned  portion  of  the  crystal 
is  revolved  on  an  axis  at  right  angles  to  this  plane, 
through  an  angle  of  180°.  That  is  to  say,  suppose 
we  made  a  clean  cut  through  the  octahedron  in  the 

130 


PLATE  XL. 

ISOMETRIC  SYSTEM. 

FIG.  1.  Spinel,  Twinned  Octahedron.         FIG.  2.  Profile. 
FIG.  3.  Shows  Positions  of  Twinning  Axis  and  Plane. 


9* 


132         HINTS  FOB  CKYSTAL  DRAWING 

direction  of  the  plane  and  twisted  the  separated 
lower  half,  half  way  round,  we  should  have  a 
twinned  octahedron. 

We  must  find  this  axis  of  rotation,  this  twinning 
axis ;  to  do  this  we  make  a  profile.  The  profile  shall 
be  taken  in  a  plane  parallel  to  the  face  {110}  of  the 
rhombdodecahedron. 

Bisect  the  angle  between  the  +  ends  of  the  al 
and  a2  axes  both  of  the  0  plan  and  C  projection  ;  the 
profile  or  section  passes  through  the  bisecting  line. 
It  lies  in  a  plane  of  symmetry  both  of  the  twinned 
and  untwinned  crystal. 

Construct  the  profile  thus  (see  Fig.  2) :  draw  a 
line  XT'  =  XP  of  the  0  plan,  the  line  bisecting  the 
angle  between  the  axes  ;  at  X'  raise  a  vertical,  make 
it  equal  the  az  axis,  join  the  ends  CC  to  Pf.  This  is 
the  profile  through  the  octahedron  at  line  XP. 

Divide  the  line  CP'  into  half  at  N.  Join  N  to 
X',  NX'  is  the  profile  of  half  the  twinning  plane,  it 
is  parallel  to  Pff.  Draw  XT'  at  right  angles  to 
NX  and  P'G ;  this  is  the  semi-profile  of  the  twinning 
axis. 

Through  point  T  where  it  meets  the  octahedral 
face,  raise  a  perpendicular  to  cut  X'P  in  S'.  Mark 
the  distance  X'S'  on  the  0  plan  in  XS.  XS  will  be 
the  0  plan  of  the  semi-twinning  axis.  Drop  S  by  a 
vertical  projector  to  the  C  projection.  Mark  on  the 
projector  downwards  from  S  of  the  C  projection  the 
distance  S'T  taken  from  the  profile  in  T.  T  will  be 
the  point  where  the  twinning  axis  meets  the  octa- 
hedral face  in  C  projection. 

Join  the  angles  1,  2,  3  of  the  octahedron  to 
point  T  producing  each  line  beyond  T,  making  TV 


TWINNED  CRYSTALS  133 

equal  71,  T2'  equal  T2,  T3'  equal  T3.  Join  points 
1',  2',  3'  to  each  other  and  to  the  points  where  the 
twinning  plane  cuts  the  edges  of  the  octahedron,  the 
figure  will  then  be  complete. 

We  can  certify  the  construction  by  this  consider- 
ation. The  face  1',  2',  3'  of  the  twinned  portion 
lies  in  the  same  plane  as  the  face  1,  2,  3  of  the 
original  octahedron.  Point  T  therefore  is  common 
to  both  faces,  is  the  centre  of  both  faces,  therefore 
the  distances  71,  T2,  T3  must  equal  the  distances 
TV,  T2',  T3f. 

This  is  one  of  the  characteristics  of  a  form  of 
which  the  experienced  crystal  draughtsman  takes 
quick  advantage  ;  the  student  should  always  be  on 
the  alert  for  such  welcome  short  cuts  through 
intricate  problems. 

The  plane  of  twinning  is  in  all  cases  a  plane  of 
symmetry  in  the  twinned  but  not  in  the  untwinned 
crystal. 

Plate  XLI.  shows  twins  from  the  tetragonal  and 
orthorhombic  systems. 

Fig.  1,  a  crystal  of  cassiterite,  c  axis  =  *67,  gives 
a  combination  of  unit  prism  {110}»  a  prism  of  the 
second  order  {100}.  a  unit  pyramid  {111}  and  a 
pyramid  of  the  second  order  {101}- 

The  twinning  plane  parallel  to  a  face  of  the 
pyramid  of  the  second  order,  is  shown  shaded  Fig.  1. 

Fig.  2  shows  the  construction  for  the  twinned 
crystal. 

Having  first  drawn  the  crystal  in  the  usual  posi- 
tion the  twinned  portion  may  be  found  thus : 
Through  point  0,  the  point  on  the  vertical  axis  cut 
by  the  twinning  plane,  draw  a  line  OU  parallel  to 


134         HINTS  FOE  CEYSTAL  DEAWING 

a  line  from  the  —  end  of  the  a2  axis  to  the  -f  end  of 
tlie  c  axis,  let  it  cut  a  vertical  through  the  centre  of 
the  prism  face  {010}  at  v.  Through  the  point  of 
cutting  draw  a  short  line  parallel  to  the  0,1  axis  to 
cut  the  vertical  edges  of  the  face  at  Zf  and  Wf. 

Through  the  point  0  draw  another  line  parallel 
to  the  #!  axis  to  cut  a  vertical  through  the  centre  of 
the  prism  face  {100}»  through*  the  point  of  cutting 
pass  a  short  line  parallel  to  OU  to  cut  the  vertical 
edges  of  the  face  {100}  at  Fand  Z. 

Repeat  for  corresponding  back  face  of  prism 
{TOO}-  Join  the  points  thus  found  Z  to  Z',  W  to 
W,  the  lines  joining  will  be  the  intersections  of 
the  twinning  plane  with  tli§  faces  of  the  prism  of 
the  first  order  {110}. 

We  next  proceed  to  find  the  c  axis  of  the 
twinned  portion  (this  must  not  be  confounded 
with  the  twinning  axis  or  axis  of  rotation  which  is 
at  right  angles  to  the  twinning  plane  but  need  not 
be  shown).  -  The  twinned  c  axis  we  find  by  a 
profile  seen  at  Fig.  3. 

This  profile  we  take  through  the  &2  and  c  axes 
because  the  vertical  plane  which  contains  them 
contains  also  the  twinned  c  axis. 

To  construct  the  profile,  first  draw  a  vertical 
line  /,  c,  this  represents  a  portion  of  the  c  axis, 
through  point  0',  the  point  where  the  c  axis  is  cut 
by  the  twinning  plane,  draw  line  0%  making  with 
the  vertical  axis  the  same  angle  9  as  the  twinning 
plane  makes  ;  to  get  this  angle  draw  the  horizontal 
line  ffh  equal  the  semi  a2  axis,  and  the  vertical  ht 
equal  the  semi  c  axis.  Join  0'  to  t,  Ot  will  be  the 
profile  of  a  portion  of  the  twinning  plane. 


136          HINTS  FOE  CKYSTAL  DEAWING 

Draw  line  O'n  making  the  angle  Ot'n  equal  cOft, 
line  O'n  will  be  the  profile  of  the  new  c  axis,  make 
On  equal  the  distance  the  apex  of  the  crystal  is 
from  the  point  where  the  twinning  plane  cuts  the 
c  axis.  The  distance  n,  I  will  be  the  plan  of  the 
portion  of  the  twinned  axis  from  O  to  n,  mark  this 
plan  distance  on  the  0  plan  from  0"  to  nf.  Drop 
n'  to  the  C  projection  at  #',  the  exact  position  of  vf 
being  obtained  by  marking  the  distance  kfl  =  ftfo  +  ol 
from  the  profile  downwards  from  the  line  Qk,  which 
is  a  production  of  the  a2  axis.  The  point  v  thus 
found  is  the  apex  of  the  twinned  portion  of  the 
crystal ;  the  c  axis  may  be  drawn  from  vf  through  0. 

Join  the  apex  of  the  crystal  to  the  apex  of  the 
twinned  portion  v'.  The  joining  line  will  be  at 
right  angles  to  the  twinning  plane,  therefore  parallel 
to  the  twinning  axis.  A  line  from  e  drawn  parallel 
to  this  joining  line  will  enable  us  to  get  point  d,— 
the  point  where  the  pyramids  replace  the  prisms  of 
the  twinned  portion — by  marking  the  distance  of  e 
from  the  twinning  plane,  along  the  line  on  the  other 
side  of  the  plane  to  d. 

Now  in  such  parallel  lines  we  have  great  aid, 
since  all  corresponding  points  of  the  two  portions 
of  the  twin  will  lie  on  such  lines  at  equal  distances 
from  the  twinning  plane. 

We  can  thus  derive  X  from  Sby  drawing  through 
S  a  short  parallel  to  e,  d  to  cut  the  central  line  of  the 
twinning  plane  produced  and  marking  the  distance 
that  S  is  to  the  left  of  the  point  of  cutting,  to  the 
right  in  X. 

Join  X  to  d  and  producing  the  line  an  equal 
distance  beyond  d  we  have  the  point  g.  The  com- 


TWINNED  CKYSTALS  137 

pletion  of  this  end  of  the  twinned  portion  will  be 
easy. 

For  the  other  end,  to  get  point  q,  the  point  on  the 
new  c  axis  where  the  pyramids  replace  the  prisms, 
we  have  only  to  draw  a  parallel  to  de  through  r 
(r  being  obtained  by  making  Qr  =  Qe)  to  cut  the 
twinned  c  axis  produced  beyond  0  in  q.  Through  q 
draw  a  line  qX'  parallel  to  dX. 

This  end  of  the  twinned  portion  can  now  be 
easily  finished  by  "  parallelism  "  and  "  repeti- 
tion ". 

Fig.  5  shows  a  twin  of  the  orthorhombic  system. 
A  crystal  of  aragonite,  axes  *66  : 1  :  '72.  It  is  a 
combination  of  the  unit  prism  {110},  brachypinacoid 
{010}  and  brachydome  {Oil}-  It  is  twinned  on  a 
plane  parallel  to  the  face  {110}-  The  twinning  plane 
is  shown  shaded  at  Fig  4. 

The  construction  offers  little  difficulty.  Draw 
the  crystal  and  Oplan.  Draw  the  line  #,/on  the  0 
plan  parallel  to  the  face  {110} ;  0,/is  the  plan  of  the 
twinning  plane.  Make  angle  0  equal  angle  6  and 
ffkf  equal  gk  and  complete  the  plan  of  the  twinned 
portion  as  shown.  Cut  the  upper  edge  of  the  dome 
by  a  projector  dropped  from  g  in  gf.  Cut  the  in- 
tersection edges  of  pinacoidal  and  domal  faces  by 
projectors  from  e  and /in  eff.  Join  e'f  to  8  for  the 
intersection  of  the  twinned  domes.  Vertical  lines 
through  d  and  /  will  be  the  intersections  of  the 
twinned  pinacoid  faces. 

There  is,  however,  one  slight  difficulty.  At  first 
sight  it  may  not  be  apparent  how  the  direction  of  the 
edge  g'Kft  the  uppermost  edge  of  the  twinned  portion 
of  the  dome  is  determined. 


138         HINTS  FOE  CKYSTAL  DBA  WING 

To  obtain  it  we  may  use  the  same  method  of 
construction  as  for  finding  the  axes. 

We  will  first  find  a  line  o,  n  through  the  centre 
of  the  crystal  in  C  projection,  parallel  to  which  g'W 
may  be  drawn. 

For  this  purpose  draw  a  horizontal  line  g,  h 
through  the  centre  of  the  0  plan.  Draw  a  vertical 
projector  from  kf  to  cut  an  H  line  oo'  drawn  through 
the  centre  of  the  C  projection,  in  point  p. 

The  vertical  projector  cuts  the  line  g,  h  in  point  I. 
Draw  the  short  line  k,  i  parallel  to  g,  k,  that  is 
parallel  to  the  a  axis. 

With  the  dividers  convey  half  the  distance  i,  I  to 
the  C  projection  and  mark  it  upwards  from  point  p 
on  the  vertical  projector  in  n,  it  will  be  the  amount 
of  apparent  elevation  the  point  K  will  assume  above 
the  horizontal,  for  since  V  lies  behind  the  plane  of 
projection  it  will  appear  elevated  not  depressed.  Join 
o  to  n,  on  is  a  horizontal  line  through  the  centre  of 
the  crystal  having  the  direction  the  upper  edge  of  the 
dome  will  take,  to  which  the  edge  of  the  dome  may 
be  drawn  parallel. 

The  figure  is  completed  by  the  usual  methods. 

Plates  XLIL  and  XLIII.  give  the  construc- 
tion for  a  twinned  -f  scalenohedron  {2131}  of  cal- 
cite,  c  axis  =  "84,  belonging  to  the  hexagonal 
system. 

It  is  twinned  about  the  -  rhombohedron  {0221}- 

Fig.  1  Plate  XLIL  is  the  crystal  drawn  in  C  pro- 
jection in  the  customary  position.  The  twinning 
plane  {0221}  is  shown  shaded. 

When  the  scalenohedron  has  been  constructed 
according  to  the  directions  given  on  Plate  XXVI. 


(2) 


PLATE  XLII. 
HEXAGONAL  SYSTEM. 

Calcite,  c_axis  =  '85. 

FIG.  1.    +  Scalenohedron  {2131},  showing  Twinning  Plane  parallel 
to  Face  of  -  Rhombohedron  {02^1}. 
FIG.  2.  Profile. 


140         HINTS  FOE  CEYSTAL  DEAWING 

we  may  find  the  intersections  of  the  twinning  plane 
thus  :  — 

First  find  the  inclination  of  the  plane.  A  line 
passing  through  the  centre  of  the  axes  and  cutting 
the  front  lower  long  edge  #,  T/,  and  the  corresponding 
upper  back  long  edge  v,  kf  in  their  centre  points  g,  h 
will  give  this  inclination. 

The  profile  shown  at  Fig.  2  will  prove  this  ;  it 
is  taken  through  the  edges  NON'  of  the  0  plan  ; 
k,  V,  Kvf  of  the  C  projection. 

This  profile  is  a  parallelogram,  formed  by  the 
edges  vk,  kv',  vK  and  &V. 

The  profile  of  the  twinning  plane  will  pass 
through  the  centre  parallel  to  the  edges  vk  and 


We  get  the  data  for  drawing  the  parallelogram 
from  the  0  plan  thus  :— 

Draw  a  vertical  line  #'V"  (see  Fig.  2),  make  the 
portion  cffd  =  the  c  axis,  make  the  line  v"v"f  =  the 
c  axis  x  3.  This  vertical  line  is  the  geometrical  axis 
of  the  scalenohedron.  Draw  a  horizontal  line 
through  the  centre  of  the  axis. 

The  0  plan  of  the  scalenohedron  is  a  hexagon, 
the  sides  of  which  are  at  right  angles  to  the  H  axes, 

Now  our  profile  is  taken  through  the  edges  ON, 
ON'. 

Marking  the  centre  of  the  profile  (7  we  convey 
the  measurements  ON,  ON'  to  the  horizontal  line 
through  O. 

Next  find  N".  This  is  obtained  by  producing 
the//  traces  of  the  planes  {2131}  and  {2311}  on 
the  0  plan.  They  meet  in  point  N".  ON"  is  there- 
fore the  intersection  edge  of  those  planes  on  the  0 


TWINNED  CEYSTALS  141 

plan.  Convey  point  N"  to  the  profile.  Join  N"  to 
0"  for  the  edge  v,  k. 

A  vertical  raised  through  point  N'  to  cut  xf'N" 
will  give  point  /,  the  point  where  the  upper  and 
lower  edges  meet ;  the  parallelogram  d'NiP'N"  can 
now  be  completed. 

The  next  step  is  to  find  the  line  cf,  the  line  of 
inclination  of  the  twinning  plane  ;  to  do  this  we 
refer  again  to  the  0  plan. 

In  accordance  with  the  symbol  {0221}  of  the 
twinning  plane  we  draw  the  line  de  on  the  0  plan, 
joining  the  J  points  of  the  -  as  and  +  a2  axes. 
de  cuts  the  line  NO  in  f.  Convey  Of  to  the  pro- 
file in  Oy,  /  joined  to  c  gives  us  the  line  of  inclina- 
tion of  the  twinning  plane  in  profile,  which  is 
parallel  to  the  edge  v"l. 

For  since  de  cuts  the  line  ON"  at  the  J  point, 
and  O'c  in  accordance  with  the  symbols  of  the  planes 
is  ^  0V',  the  triangles  O'cf  and  O'tf'N"  are  similar. 

The  line  g,  h'  drawn  parallel  to  the  line  of  in- 
clination is  the  profile  of  the  twinning  plane  in 
position  and  since  it  passes  through  the  centre  of 
the  parallelogram  JYYWV",  parallel  to  the  sides, 
it  must  bisect  the  sides  Nv"  and  N"vm. 

Having  ascertained  their  position  on  the  edges 
of  the  scalenohedron,  from  the  profile  we  can  at 
once  mark  the  two  points  g,  h  in  C  projection. 

We  have  to  find  six  other  points  where  the 
twinning  plane  meets  the  edges.  We  have  two 
more  in  the  ends  of  the  a\  axis,  since  the  plane 
passes  through  that  axis.  The  other  four  points 
must  lie  somewhere  on  the  edges  of  the  scaleno- 
hedron between  the  points  already  obtained. 


142         HINTS  FOE  CEYSTAL  DEAWING 

They  are  easily  found.  Take  the  one  on  the  edge 
u,  v,  we  get  it  thus  :  join  the  +  end  of  the  al  axis 
to  g,  bisect  the  joining  line  and  draw  through  the 
centre  of  the  scalenohedron  and  the  bisection  a 
line,  producing  it  to  cut  u,  v  in  1  ;  1  is  another  of 
the  required  points,  the  others  can  be  found  by 
"  parallelism  "  ;  a  line  from  1  parallel  to  the  a^  axis 
will  cut  the  edge  corresponding  to  u,  v  in  2,  etc. 

The  explanation  of  the  method  used  for  finding 
point  1  is  as  follows  : — 

Point  i  on  the  0  plan  is  the  plan  of  g,  the  point 
where  the  twinning  plane  meets  the  edge  rkf.  Join 
i  to  the  +  end  of  the  a^  axis.  The  bisection  of  i,  + 
#!,  point  S',  falls  somewhere  vertically  below  the 
edge  MO,  i.e.,  u,  v  in  C  projection.  If  now  we 
imagine  a  vertical  plane  passed  through  that  edge 
it  will  contain  the  centre  point  0,  the  point  of  bi- 
section S,  and  consequently  a  line  passed  through 
the  points  0  and  S,  which  line  will  also  lie  in  the 
twinning  plane,  since  0  and  S  are  in  that  plane  ; 
clearly  then  where  such  a  line  meets  the  edge  will  be 
a  point  where  the  twinning  plane  cuts  the  edge  as 
SJiown  at  point  1  in  C  projection. 

To  construct  the  crystal  twinned  about  this 
plane  is  the  next  problem. 

But  this  position  of  the  scalenohedron  is  not 
suitable  for  showing  the  twinned  form  well.  It  will 
be  better,  to  rotate  the  crystal  a  little  to  the  right 
until  the  a\  axis  takes  the  position  of  the  a^  axis  of 
the  isometric  system  (Plate  XLIII.  Fig.  1). 

In  Fig.  1  this  has  been  done  by  putting  the  0 
plan  with  the  a^  axis  in  the  desired  position. 

Draw   the    C  projection   in   the   new  position, 


-ft, 


PLATE  XLIII. 
HEXAGONAL  SYSTEM. 
Calcite  c  axis  =  '85. 

FIG.  1.  +  Scalenohedron  {2131},  Twinning  Plane  -  Rhombohedron 
{0221}. 

FIG.  2.  Profile  for  finding  c  axis  of  Twinned  Portion. 


144          HINTS  FOE  CKYSTAL  DEAWING 

leaving  the  lower  half  in  pencil.  Find  the  inter- 
sections of  the  twinning  plane,  lines  1-2,  3-4,  5-6,  7-8, 
by  the  method  explained  for  Plate  XLII.  Fig.  1. 

The  next  step  is  to  find  the  twinned  axis  by 
the  profile  shown  at  Fig.  2.  Line  m'n  is  the  profile 
of  the  twinning  plane,  got  by  making  O'mf  equal 
Om  of  the  0  plan  and  O'O"  equal  the  semi-height 
of  the  c  axis.  At  0"  construct  the  angle  ff,  equal 
0,  that  is  to  say,  equal  to  the  angle  the  vertical  axis 
makes  with  the  twinning  plane ;  the  line  O"Z"  will 
be  the  position  of  the  twinned  axis. 

Make  0"Z"  =  thrice  the  true  semi-length  of  the 
c  axis,  that  is  to  say,  the  true  semi-height  of  the 
scalenohedron. 

Through  0"  draw  the  horizontal  OnX'  and  from 
Z"  drop  a  vertical  to  cut  it ;  the  angle  ft  will  be  the 
true  angle,  the  new  axis  makes  with  the  imaginary 
horizontal  plane,  through  the  H  axes ;  the  height 
X'Z"will  be  the  trueheight  of  the  apex  of  the  twinned 
portion  above  that  horizontal  plane. 

Now  on  the  0  plan  produce  the  line  ON  to  Z 
making  OZ  equal  0"X'  the  plan  length  of  the  new 
axis  obtained  by  the  profile. 

Draw  thje  line  OX  on  the  C  projection,  bisecting 
the  angle  made  by  the  +  08  and  -  a2  axes.  Drop  a 
vertical  projector  from  Zto  cut  this  line  in  X.  Mark 
the  height  X'Zt'  on  the  projector  in  Z'.  Z'  will  be 
the  apex  of  the  twinned  portion  of  the  crystal  in  C 
projection. 

To  the  apex  thus  found  we  can  at  once  join  the 
points  2,  3,  4. 

Carry  the  new  axis  beyond  0  to  the  right  and 
mark  off  the  distance  OZ"'  from  the  other  end. 


TWINNED  CKYSTALS  145 

To  find  point  K  of  the  twinned  portion,  join  the 
apex  Z  of  the  twinned  portion  to  the  apex  V  of  the 
other  part ;  then  from  point  k  draw  a  line  parallel 
to  that  joining  the  apices  and  make  the  distance 
from  K  to  q,  the  point  where  this  parallel  cuts  the 
twinning  plane,  equal  qk.  Join  kf  to  point  7. 

To  get  points  I  and  /',  through  points  6  and  8, 
the  points  where  the  twinning  plane  meets  scale- 
nohedron  edges,  draw  lines  from  the  apex  Zf  to  meet 
lines  drawn  from  i  and  i  parallel  to  that  joining  the 
apices. 

The  figure  is  now  complete. 

Plate  XLIV.  shows  twinning  in  the  monoclinic 
and  triclinic  systems.  Fig.  1  is  a  crystal  of  gypsum, 
axes  *69 :  1  :  *4,  ft  angle  84°  42',  with  the  twinning 
plane  parallel  to  the  vertical  axis.  The  plane  is 
shown  shaded. 

Fig.  2  shows  the  twinned  crystal,  composed  of 
unit  prism  {110}.  clinopinacoid  {010}  and  unit 
pyramid  {111}- 

Find  the  axes  according  to  the  construction  given 
for  the  monoclinic  system  (see  Plate  XXXIII.), 
showing  the  position  of  point  e,  the  level  of  the  + 
end  of  the  a  axis  if  it  were  not  inclined. 

Draw  the  crystal  in  pencil,  finding  the  points  g, 
h,  i,  I  through  which  the  vertical  intersections  of  the 
prism  {110}  and  the  clinopinacoid  {010}  pass. 

The  front  portion  is  the  twinned  portion.  From 
point/,  where  the  front  unit  pyramid  edge  meets  the 
back  edge  of  the  prism,  draw  a  line  parallel  not  to  the 
a  axis  Oa,  but  to  line  Oe,  the  position  of  the  axis  if  it 
were  horizontal.  Where  this  line  cuts  the  front  edge 

of  the  prism  in  f  will  be  the  twinned  position  of  f. 

10 


146          HINTS  FOE  CEYSTAL  DEAWING 

Lines  parallel  to  ff  from  /?,  p,  m  will  find  the 
corresponding  twinned  points  n'  p'  m'.  The  edges 
may  then  be  drawn  by  "  parallelism  ". 

Fig.  4  is  a  triclinic  crystal,  labradorite,  axial  ratio 
•63  : 1 :  '55,  angles  a  =  93°  23',  £  =  116°  29',  y  =  89°  59'. 

The  twinning  plane  shown  shaded  is  {010}, 
parallel  to  the  brachypinacoid. 

The  figure  is  a  combination  of  brachypinacoid 
{010},  base  {001}»  macrodome  {201}  and  unit  prism 
{110}. 

Having  found  the  axes  according  to  the  construc- 
tion for  the  triclinic  system  (see  Plate  XXXVII.), 
the  intersection  d,  e  of  unit  prism  {110}  and  base  is 
drawn  parallel  to  the  line  joining  the  +  end  of  the  a 
and  —  end  of  the  b  axes. 

The  intersection  of  prism  and  pinacoid  d,  f  is  a 
vertical  line.  Joining  e  to  /,  forms  the  intersection 
edge  between  macrodome  {201}  and  prism,  which 
if  correct  will  be  parallel  to  line  k,  k',  the  intersection 
of  the  planes  illO}  and  {20l Mound  by  drawing  the 
line  n,  k  parallel  to  the  b  axis  through  the  |  point  of 
the  +  a  axis,  to  cut  line  a,  b,  from  the  +  end  of  the 
a  to  the  —  end  of  the  b  axes  in  k. 

k,  n  and  a,  b  are  the  basal  traces  respectively  of 
the  planes  {ll0}  and  {201}-  kf  is  found  by  drawing 
b,  kf  a  vertical  through  the  —  end  of  the  b  axis  to  cut 
kfn,  a  line  parallel  to  the  b  axis  through  the  —  end 
of  the  c  axis. 

These  last  lines  are  respectively  the  vertical 
traces  of  the  planes  ilTO}  and  {201}  on  the  axial 
plane  through  the  b  and  c  axes ;  k,  kf  is  their  inter- 
section. 

The  twinned  portion  is  simple  of  construction. 


PLATE  XLIV. 

MONOCLINIC    AND    TRICLINIC    SYSTEMS. 

FIG.  1.  Gypsum.  Axial  Ratio  -69 : 1 :  '4 ;  /3  80°  42' ;  showing 
Twinning  Plane,  Orthopinacoid  {100}. 

Fic,T  2.  Twinned  Crystal. 

FIG.  3.  Labradorite.  Axial  Ratio  '63 : 1 :  "55  ;  a  93°  23' ;  /3  116°  29' ; 
y  89°  59' ;  showing  Twinning  Plane,  Pinacoid  {010. 

FIG.  4.  Twinned  Crystal. 


10 


148          HINTS  FOR  CRYSTAL  DRAWING 

Point  i't  the  corresponding  point  to  i,  is  found  by 
drawing  the  line  i,  if  parallel  not  to  the  b  axis  but  to 
line  01,  the  position  it  would  take  if  not  inclined. 
All  similarly  corresponding  points  are  found  in  like 
manner,  as  d'f's'. 

If  the  student  has  had  the  patient  perseverance 
to  work  out  the  construction  and  follow  the  reason- 
ing in  all  the  problems  on  crystal  forms  here 
given,  he  should  be  in  a  position— with  the  aid  of 
careful  thought— to  draw  in  C  projection  any  form 
or  combination  of  forms,  he,  as  a  mineralogist,  may 
wish  to  represent,  supposing  he  can  obtain  the 
necessary  data  forangles  and  measurements. 

On  the  other  hand,  should  he  only  wish  to 
follow  the  subject  of  crystal  projection  far  enough 
to  enable  him  to  pass  school  examinations,  he  need 
only  concern  himself  with  the  more  common  forms, 
the  constructions  for  which,  he  may  easily  master 
and  commit  to  memory. 

Exercises — Problems  for  solution.  Twins. 
Isometric  system — Twinned  Pyritohedron. 
Tetragonal  system— Rutile  c  axis  =  '64;  {110}; 

{1001;  {111},  twinned  on  {101}- 
Tetragonal  system— Rutile,  {110};  {1001;  {111}; 

{101},  twinned  on  {301}. 
Hexagonal  system— Calcite  c  axis  =  '85  ;  {10111, 

twinned  on  {0112}. 
Monoclinic    system -- Hornblende,  axial  ratio 

•05  : 1  :  -29  ;  /3  73°  58' ;  {1 10} ;  {010! ;   {01 1 } ; 

{101},  twinned  on  {100}- 


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